Step 1: Establishing the Function and Derivative

Given the function $f(x) = x^2 - x + 1$. To determine the intervals of increase and decrease, we must calculate the first derivative with respect to $x$ [By the Power Rule of Differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$]:

$f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(x) + \frac{d}{dx}(1)$

$f'(x) = 2x - 1$

Step 2: Finding Critical Points

A function changes its monotonicity behavior at critical points where $f'(x) = 0$.

$2x - 1 = 0$

$2x = 1 \implies x = \frac{1}{2}$

Since $\frac{1}{2}$ lies within the interval $(-1, 1)$, we must evaluate the behavior of $f'(x)$ on either side of this critical point within the given domain.

Step 3: Interval Analysis

We partition the interval $(-1, 1)$ into two sub-intervals based on the critical point $x = \frac{1}{2}$:

Step 4: Logical Deduction

Per the Monotonicity Theorem, a function $f$ is strictly increasing on an interval if $f'(x) > 0$ for all $x$ in that interval, and strictly decreasing if $f'(x) < 0$ for all $x$ in that interval.

As demonstrated in Step 3:

• For $x \in (-1, 1/2)$, $f'(x) < 0$, therefore $f$ is strictly decreasing.
• For $x \in (1/2, 1)$, $f'(x) > 0$, therefore $f$ is strictly increasing.

Since the derivative changes sign within the interval $(-1, 1)$, the function fails the criteria for being strictly monotonic throughout the entire domain.

Final Solution: Because the derivative $f'(x) = 2x - 1$ takes both negative and positive values in the interval $(-1, 1)$, the function is neither strictly increasing nor strictly decreasing on the given interval.

Step 1: Definition of the Logarithmic Function

Let $f(x)$ be the natural logarithmic function defined on the domain $(0, \infty)$. The function is represented as:

$f(x) = \ln(x)$

Step 2: Differentiation of the Function

To determine the interval of increase, we must calculate the first derivative of the function with respect to $x$.

$f'(x) = \frac{d}{dx}[\ln(x)]$

[By the standard derivative rule for natural logarithms: $\frac{d}{dx}(\ln x) = \frac{1}{x}$]

$f'(x) = \frac{1}{x}$

Step 3: Analyzing the Sign of the Derivative

According to the First Derivative Test for monotonic functions, a function $f$ is strictly increasing on an interval if $f'(x) > 0$ for all $x$ within that interval.

• We are given the domain $x \in (0, \infty)$.
• For all $x > 0$, the value of $\frac{1}{x}$ is always positive.
• Mathematically, if $x > 0$, then $\frac{1}{x} > 0$.

[By the properties of real numbers and division, the reciprocal of a positive number is strictly positive.]

Step 4: Geometric Visualization

Step 5: Conclusion based on Derivative Test

Since $f'(x) = \frac{1}{x} > 0$ for every $x \in (0, \infty)$, the slope of the tangent to the curve is always positive. Therefore, the function $f(x) = \ln(x)$ is strictly increasing on its entire domain $(0, \infty)$.

Final Solution: Since $f'(x) = \frac{1}{x} > 0$ for all $x \in (0, \infty)$, the logarithmic function is strictly increasing on the interval $(0, \infty)$.

Step 1: Establishing the Function and Derivative

Given the function $f(x) = 2x^2 - 3x$, we proceed by finding the first derivative with respect to $x$, which represents the rate of change of the function [Per the Power Rule of Differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$].

$f'(x) = \frac{d}{dx}(2x^2 - 3x) = 4x - 3$

Step 2: Determining Critical Points

To find the intervals of increase or decrease, we identify the critical points by setting the first derivative equal to zero [By Fermat’s Theorem on stationary points].

$4x - 3 = 0$

$x = \frac{3}{4}$

Step 3: Interval Analysis

The critical point $x = \frac{3}{4}$ divides the real number line into two distinct intervals: $(-\infty, \frac{3}{4})$ and $(\frac{3}{4}, \infty)$. We analyze the sign of $f'(x)$ in each interval to determine monotonicity [By the First Derivative Test for Monotonicity].

Step 4: Logical Deduction of Increasing Interval

A function $f$ is strictly increasing on an interval if $f'(x) > 0$ for all $x$ within that interval. Based on the analysis in Step 3, $f'(x) > 0$ when $x \in (\frac{3}{4}, \infty)$.

Final Solution: The function $f(x) = 2x^2 - 3x$ is increasing on the interval $(\frac{3}{4}, \infty)$.

Step 1: Establishing the Derivative

Given the function $f(x) = x^2 + ax + 1$, we must first determine the condition under which the function is increasing on the interval $[1, 2]$. A function $f(x)$ is increasing on an interval if its first derivative, $f'(x)$, is greater than or equal to zero for all $x$ in that interval [Per the First Derivative Test for Monotonicity].

Calculating the derivative with respect to $x$:

$f'(x) = \frac{d}{dx}(x^2 + ax + 1)$

$f'(x) = 2x + a$

Step 2: Defining the Condition for Monotonicity

For $f(x)$ to be increasing on $[1, 2]$, we require:

$f'(x) \geq 0$ for all $x \in [1, 2]$

$2x + a \geq 0$

$a \geq -2x$

Step 3: Determining the Range of $a$ via Interval Analysis

Since the inequality $a \geq -2x$ must hold for all $x$ in the interval $[1, 2]$, the value of $a$ must be greater than or equal to the maximum possible value of the expression $-2x$ on the interval $[1, 2]$.

Let $g(x) = -2x$. Since $g(x)$ is a strictly decreasing function (as the coefficient of $x$ is negative), its maximum value on the closed interval $[1, 2]$ occurs at the lower bound, $x = 1$.

Calculating the maximum value of $-2x$ on $[1, 2]$:

$g(1) = -2(1) = -2$

Thus, for $a \geq -2x$ to hold true for all $x \in [1, 2]$, $a$ must satisfy:

$a \geq -2$

Step 4: Geometric Representation of the Derivative

Step 5: Logical Synthesis

By evaluating the inequality at the extreme points of the domain, we ensure the derivative remains non-negative throughout the entire specified interval. If $a \geq -2$, then $2x + a \geq 2(1) + (-2) = 0$ for all $x \geq 1$. Therefore, the condition $f'(x) \geq 0$ is satisfied for all $x \in [1, 2]$.

Final Solution: $a \in [-2, \infty)$

Step 1: Given Function and Derivative Calculation

Let the given function be $f(x) = -2x^3 - 9x^2 - 12x + 1$.

To determine the intervals of strictly increasing or decreasing behavior, we must find the first derivative of $f(x)$ with respect to $x$ [Per the Derivative Power Rule, $\frac{d}{dx}(ax^n) = anx^{n-1}$]:

$f'(x) = \frac{d}{dx}(-2x^3 - 9x^2 - 12x + 1)$

$f'(x) = -6x^2 - 18x - 12$

Step 2: Determining Critical Points

Critical points occur where $f'(x) = 0$ or where $f'(x)$ is undefined. Since $f'(x)$ is a polynomial, it is defined everywhere. We solve for $f'(x) = 0$:

$-6x^2 - 18x - 12 = 0$

Divide the entire equation by $-6$ [Algebraic simplification]:

$x^2 + 3x + 2 = 0$

Factor the quadratic equation:

$(x + 1)(x + 2) = 0$

Thus, the critical points are $x = -1$ and $x = -2$.

Step 3: Interval Analysis

The critical points divide the real number line into three distinct intervals: $(-\infty, -2)$, $(-2, -1)$, and $(-1, \infty)$. We test the sign of $f'(x) = -6(x+1)(x+2)$ in each interval:

Visual Representation

Step 4: Conclusion based on the First Derivative Test

• [Per the First Derivative Test for Monotonicity]: If $f'(x) > 0$, the function is strictly increasing; if $f'(x) < 0$, the function is strictly decreasing.
• The function is strictly increasing on the interval $(-2, -1)$.
• The function is strictly decreasing on the intervals $(-\infty, -2) \cup (-1, \infty)$.

Final Solution: The function is strictly increasing in $(-2, -1)$ and strictly decreasing in $(-\infty, -2) \cup (-1, \infty)$.

Step 1: Establishing the Derivative Function

To determine the intervals of monotonicity for the function $f(x) = \sin x$, we must examine the sign of its first derivative, $f'(x)$, with respect to $x$.

Given: $f(x) = \sin x$

Differentiating with respect to $x$:
$f'(x) = \frac{d}{dx}(\sin x) = \cos x$ [By the derivative rule for trigonometric functions]

Step 2: Analyzing the Critical Points in the Interval $(0, \pi)$

A function is strictly increasing if $f'(x) > 0$ and strictly decreasing if $f'(x) < 0$. To analyze the behavior of $f'(x) = \cos x$ in the interval $(0, \pi)$, we must identify the point where the derivative changes sign.

Setting $f'(x) = 0$:
$\cos x = 0$
In the interval $(0, \pi)$, the only solution is $x = \frac{\pi}{2}$.

Step 3: Evaluating Monotonicity via Sub-intervals

We partition the interval $(0, \pi)$ into two sub-intervals based on the critical point $x = \frac{\pi}{2}$:

Step 4: Geometric Visualization

Step 5: Synthesis of Results

Per the definition of monotonic functions [Definition of Monotonicity], a function is called increasing on an interval $I$ if $f'(x) \geq 0$ for all $x \in I$. Conversely, it is decreasing if $f'(x) \leq 0$ for all $x \in I$.

• Since $f'(x) > 0$ on $(0, \frac{\pi}{2})$, the function is strictly increasing on this sub-interval.
• Since $f'(x) < 0$ on $(\frac{\pi}{2}, \pi)$, the function is strictly decreasing on this sub-interval.

Because the derivative $f'(x)$ changes sign from positive to negative at $x = \frac{\pi}{2}$, the function does not maintain a consistent monotonic behavior across the entire interval $(0, \pi)$. Therefore, the function cannot be classified as globally increasing or globally decreasing on the specified domain.

Final Solution: Since the derivative $f'(x) = \cos x$ is positive in $(0, \pi/2)$ and negative in $(\pi/2, \pi)$, the function $f(x) = \sin x$ is neither increasing nor decreasing in $(0, \pi)$.

Step 1: Definition of the Function

Let the given function be denoted as $f(x)$:

$f(x) = x^2 + 2x - 5$

The domain of this polynomial function is $x \in \mathbb{R}$ [Per the property of polynomial functions].

Step 2: Differentiation of the Function

To determine the intervals of increase and decrease, we must find the first derivative, $f'(x)$, with respect to $x$ [By the First Derivative Test for Monotonicity].

$f'(x) = \frac{d}{dx}(x^2 + 2x - 5)$

$f'(x) = 2x + 2$

Step 3: Finding Critical Points

Critical points occur where $f'(x) = 0$ or where the derivative is undefined. Since $f'(x) = 2x + 2$ is a linear polynomial, it is defined for all $x \in \mathbb{R}$.

$2x + 2 = 0$

$2x = -2$

$x = -1$

The critical point $x = -1$ divides the real number line into two disjoint intervals: $(-\infty, -1)$ and $(-1, \infty)$.

Step 4: Sign Analysis of $f'(x)$

Step 5: Visual Representation of the Derivative Sign

Step 6: Final Deduction

• Since $f'(x) < 0$ for all $x \in (-\infty, -1)$, the function is strictly decreasing on the interval $(-\infty, -1)$.
• Since $f'(x) > 0$ for all $x \in (-1, \infty)$, the function is strictly increasing on the interval $(-1, \infty)$.

Final Solution: The function $f(x) = x^2 + 2x - 5$ is strictly increasing on $(-1, \infty)$ and strictly decreasing on $(-\infty, -1)$.

Step 1: Establishing the Derivative of the Function

Given the function $f(x) = \log(\sin x)$, we determine the interval of increase and decrease by finding the first derivative with respect to $x$ using the Chain Rule [By Chain Rule: $\frac{d}{dx} \log(u) = \frac{1}{u} \cdot \frac{du}{dx}$]:

$f'(x) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x)$

$f'(x) = \frac{\cos x}{\sin x}$

$f'(x) = \cot x$

Step 2: Analyzing the Sign of $f'(x)$ on Given Intervals

To determine if the function is increasing or decreasing, we evaluate the sign of $f'(x) = \cot x$ across the specified domains [Per the First Derivative Test for Monotonicity]:

• For the interval $(0, \frac{\pi}{2})$: In the first quadrant, all trigonometric functions are positive. Therefore, $\cot x > 0$ for all $x \in (0, \frac{\pi}{2})$. Since $f'(x) > 0$, the function $f(x)$ is strictly increasing in this interval.
• For the interval $(\frac{\pi}{2}, \pi)$: In the second quadrant, $\cos x > 0$ and $\sin x > 0$, but the ratio $\frac{\cos x}{\sin x}$ is negative. Therefore, $\cot x < 0$ for all $x \in (\frac{\pi}{2}, \pi)$. Since $f'(x) < 0$, the function $f(x)$ is strictly decreasing in this interval.

Visual Representation of $f'(x) = \cot x$

Step 3: Logical Synthesis

Final Solution: Since the derivative $f'(x) = \cot x$ is positive on $(0, \frac{\pi}{2})$, $f(x)$ is strictly increasing. Conversely, since $f'(x) = \cot x$ is negative on $(\frac{\pi}{2}, \pi)$, $f(x)$ is strictly decreasing. The proposition is proved.

Step 1: Define the Function and Its Domain

The given function is $f(x) = x + \frac{1}{x}$. The domain of this function is all real numbers except $x = 0$, denoted as $D = \mathbb{R} \setminus \{0\}$. The interval $I$ is given as being disjoint from $[-1, 1]$, which implies $I \subseteq (-\infty, -1) \cup (1, \infty)$.

Step 2: Compute the First Derivative

To determine the monotonic behavior of the function, we calculate the derivative $f'(x)$ with respect to $x$ [By the Sum Rule and Power Rule of differentiation]:

$f(x) = x + x^{-1}$

$f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1})$

$f'(x) = 1 - x^{-2} = 1 - \frac{1}{x^2}$

$f'(x) = \frac{x^2 - 1}{x^2}$

Step 3: Analyze the Sign of the Derivative on Interval $I$

For a function to be strictly increasing on an interval $I$, it is sufficient that $f'(x) > 0$ for all $x \in I$ [By the First Derivative Test for Monotonicity].

• Since $x^2$ is always positive for any $x \neq 0$, the sign of $f'(x)$ depends solely on the numerator $x^2 - 1$.
• Given that $I$ is disjoint from $[-1, 1]$, for any $x \in I$, we have $|x| > 1$.
• If $|x| > 1$, then $x^2 > 1$.
• Consequently, $x^2 - 1 > 0$.

Step 4: Logical Conclusion of Monotonicity

Since $x^2 - 1 > 0$ and $x^2 > 0$ for all $x \in I$, it follows that:

$f'(x) = \frac{x^2 - 1}{x^2} > 0$ for all $x \in I$.

Because the derivative $f'(x)$ is strictly positive on the interval $I$, the function $f(x)$ is strictly increasing on $I$ [Per the Theorem on Monotonic Functions].

Visual Representation of $f'(x)$ Sign Change

Final Solution: Since $f'(x) = \frac{x^2 - 1}{x^2} > 0$ for all $x \in I$ where $x^2 > 1$, the function $f(x) = x + \frac{1}{x}$ is strictly increasing on $I$.

Step 1: Definition of the Function

Let the given function be denoted by $f(x)$:

$f(x) = 10 - 6x - 2x^2$

Step 2: Determining the First Derivative

To find the intervals of monotonicity, we must calculate the first derivative $f'(x)$ with respect to $x$ [Per the Power Rule of differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$]:

$f'(x) = \frac{d}{dx}(10) - \frac{d}{dx}(6x) - \frac{d}{dx}(2x^2)$

$f'(x) = 0 - 6 - 4x$

$f'(x) = -2(3 + 2x)$

Step 3: Finding Critical Points

Critical points occur where $f'(x) = 0$ [By Fermat's Theorem on stationary points]:

$-2(3 + 2x) = 0$

$3 + 2x = 0$

$x = -\frac{3}{2}$ or $x = -1.5$

Step 4: Interval Analysis

The point $x = -1.5$ divides the real number line into two distinct intervals: $(-\infty, -1.5)$ and $(-1.5, \infty)$. We test the sign of $f'(x)$ in each interval:

Step 5: Visual Representation of the Monotonicity

Step 6: Conclusion

Based on the sign convention [Per the First Derivative Test for monotonicity]:

• The function $f(x)$ is strictly increasing in the interval $(-\infty, -1.5)$.
• The function $f(x)$ is strictly decreasing in the interval $(-1.5, \infty)$.

Final Solution: The function is strictly increasing on $(-\infty, -1.5)$ and strictly decreasing on $(-1.5, \infty)$.

Step 1: Defining the Function and its Derivative

Given the function: $f(x) = [x(x - 2)]^2$

Expanding the interior term first simplifies the differentiation process: $f(x) = (x^2 - 2x)^2$ [Algebraic Expansion]

Now, calculate the first derivative $f'(x)$ using the Chain Rule: $\frac{d}{dx}[u^n] = n \cdot u^{n-1} \cdot \frac{du}{dx}$ [Chain Rule of Differentiation]

$f'(x) = 2(x^2 - 2x) \cdot \frac{d}{dx}(x^2 - 2x)$

$f'(x) = 2(x^2 - 2x)(2x - 2)$

$f'(x) = 4(x)(x - 2)(x - 1)$

Step 2: Finding Critical Points

To determine the intervals of increase, we set $f'(x) = 0$ to find the critical points:

$4x(x - 1)(x - 2) = 0$

The critical points are $x = 0$, $x = 1$, and $x = 2$.

Step 3: Interval Analysis

The critical points divide the number line into four distinct intervals. We test the sign of $f'(x)$ in each interval:

Step 4: Visualizing the Function Behavior

Step 5: Conclusion

A function is strictly increasing where $f'(x) > 0$ and increasing where $f'(x) \geq 0$. Based on the interval analysis, $f'(x) > 0$ for $x \in (0, 1) \cup (2, \infty)$.

Final Solution: The function $y = [x(x - 2)]^2$ is an increasing function for $x \in [0, 1] \cup [2, \infty)$.

Step 1: Given Function and Domain

We are given the cubic polynomial function defined over the set of all real numbers $\mathbb{R}$:

$f(x) = x^3 - 3x^2 + 3x - 100$

[Per the definition of a strictly increasing function, $f(x)$ is increasing on $\mathbb{R}$ if $f'(x) \ge 0$ for all $x \in \mathbb{R}$.]

Step 2: Differentiation with respect to $x$

Applying the power rule of differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$, we calculate the first derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(3x^2) + \frac{d}{dx}(3x) - \frac{d}{dx}(100)$

$f'(x) = 3x^2 - 3(2x) + 3(1) - 0$

$f'(x) = 3x^2 - 6x + 3$

Step 3: Algebraic Factorization

To analyze the sign of $f'(x)$, we factor the quadratic expression:

$f'(x) = 3(x^2 - 2x + 1)$

[Utilizing the algebraic identity $(a - b)^2 = a^2 - 2ab + b^2$ where $a=x$ and $b=1$]:

$f'(x) = 3(x - 1)^2$

Step 4: Analysis of the Derivative

We examine the term $(x - 1)^2$ for all $x \in \mathbb{R}$:

• For any real number $x$, $(x - 1)$ is real.
• The square of any real number is always non-negative: $(x - 1)^2 \ge 0$ [Property of Squares in $\mathbb{R}$].
• Multiplying by the positive constant $3$ maintains the inequality: $3(x - 1)^2 \ge 0$.

Therefore, $f'(x) \ge 0$ for all $x \in \mathbb{R}$.

Step 5: Visual Representation of Derivative Behavior

Step 6: Conclusion based on Monotonicity

Since $f'(x) = 3(x-1)^2 \ge 0$ for all values of $x$ in the domain $\mathbb{R}$, and the derivative only equals zero at the isolated point $x = 1$ (it does not vanish on any open interval), the function $f(x)$ is strictly increasing on $\mathbb{R}$.

Final Solution: The function $f(x) = x^3 - 3x^2 + 3x - 100$ is strictly increasing in $\mathbb{R}$ because its derivative $f'(x) = 3(x-1)^2$ is non-negative for all $x \in \mathbb{R}$.

Step 1: Establishing the Function and Domain

The given function is defined as:

$f(x) = \log(1 + x) - \frac{2x}{2 + x}$

Domain: $x > -1$. [Given condition].

Step 2: Differentiating the Function

To determine the nature of the function (increasing or decreasing), we must find its first derivative $f'(x)$ with respect to $x$. [Per the Monotonicity Theorem, a function is increasing on an interval if $f'(x) \geq 0$].

$f'(x) = \frac{d}{dx} \left[ \log(1 + x) \right] - \frac{d}{dx} \left[ \frac{2x}{2 + x} \right]$

Using the chain rule for the logarithmic term: $\frac{d}{dx} \log(1+x) = \frac{1}{1+x}$.

Using the quotient rule for the rational term: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$.

Let $u = 2x \implies u' = 2$ and $v = 2+x \implies v' = 1$.

$\frac{d}{dx} \left( \frac{2x}{2+x} \right) = \frac{(2)(2+x) - (2x)(1)}{(2+x)^2} = \frac{4 + 2x - 2x}{(2+x)^2} = \frac{4}{(2+x)^2}$.

Step 3: Simplifying the Derivative

$f'(x) = \frac{1}{1+x} - \frac{4}{(2+x)^2}$

To simplify, find a common denominator:

$f'(x) = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2}$

$f'(x) = \frac{4 + 4x + x^2 - 4 - 4x}{(1+x)(2+x)^2}$

$f'(x) = \frac{x^2}{(1+x)(2+x)^2}$

Step 4: Analyzing the Sign of $f'(x)$

We analyze the expression $\frac{x^2}{(1+x)(2+x)^2}$ for the interval $x > -1$:

• $x^2 \geq 0$ for all $x \in \mathbb{R}$. [Property of squares].
• $(1+x) > 0$ since $x > -1$. [Given condition].
• $(2+x)^2 > 0$ for all $x > -1$. [Square of a real number is non-negative, and here $2+x \neq 0$].

Since the numerator $x^2 \geq 0$ and the denominator $(1+x)(2+x)^2 > 0$ for all $x > -1$, it follows that $f'(x) \geq 0$ for all $x > -1$.

Conclusion

Because $f'(x) \geq 0$ for all $x$ in the domain $(-1, \infty)$, the function $f(x)$ is an increasing function.

Final Solution: Since $f'(x) = \frac{x^2}{(1+x)(2+x)^2} \geq 0$ for all $x > -1$, the function $y = \log(1 + x) - \frac{2x}{2+x}$ is an increasing function of $x$ throughout its domain.

Step 1: Theoretical Foundation

To determine the nature of the function $f(x)$ on its domain $\mathbb{R}$, we utilize the First Derivative Test for monotonicity. According to the theorem on monotonicity: A differentiable function $f$ is strictly increasing on an interval $I$ if $f'(x) > 0$ for all $x \in I$.

Step 2: Differentiation of the Function

Given the function:
$f(x) = 3x + 17$
To find the rate of change with respect to $x$, we compute the derivative $f'(x)$ using the power rule $\frac{d}{dx}(ax^n) = anx^{n-1}$ and the constant rule $\frac{d}{dx}(c) = 0$ [Per the Sum Rule of Derivatives]:

$f'(x) = \frac{d}{dx}(3x) + \frac{d}{dx}(17)$
$f'(x) = 3(1) + 0$
$f'(x) = 3$

Step 3: Analytical Inference

Since $f'(x) = 3$, it is observed that $3 > 0$ for all values of $x$ in the domain $\mathbb{R}$.
Mathematically, $\forall x \in \mathbb{R}, f'(x) > 0$. [By the definition of an increasing function].

Step 4: Visual Representation

Conclusion

Because the derivative of the function $f(x) = 3x + 17$ is a positive constant ($3$) for all $x \in \mathbb{R}$, the function is strictly increasing on the entire set of real numbers.

Final Solution: Since $f'(x) = 3 > 0$ for all $x \in \mathbb{R}$, the function $f(x) = 3x + 17$ is strictly increasing on $\mathbb{R}$.

Step 1: Establishing the Function and Derivative

Given the function: $f(x) = x^2 e^{-x}$

To determine the intervals of increase, we must calculate the first derivative $f'(x)$ with respect to $x$ using the Product Rule, $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.

Let $u = x^2$ and $v = e^{-x}$. Then $u' = 2x$ and $v' = -e^{-x}$ [Per the Chain Rule].

$f'(x) = (2x)(e^{-x}) + (x^2)(-e^{-x})$

$f'(x) = e^{-x}(2x - x^2)$

$f'(x) = x(2 - x)e^{-x}$

Step 2: Identifying Critical Points

The function is increasing when $f'(x) > 0$. We find the critical points by setting $f'(x) = 0$:

$x(2 - x)e^{-x} = 0$

Since $e^{-x}$ is strictly positive for all $x \in \mathbb{R}$ [Property of Exponential Functions], the roots are determined solely by the polynomial component:

$x = 0$ or $2 - x = 0 \implies x = 2$.

The critical points are $x = 0$ and $x = 2$.

Step 3: Interval Analysis

We divide the number line into three intervals: $(-\infty, 0)$, $(0, 2)$, and $(2, \infty)$.

Step 4: Geometric Visualization of the Derivative Sign

Step 5: Conclusion based on Monotonicity

By the First Derivative Test for monotonicity, a function $f$ is strictly increasing on an interval if $f'(x) > 0$ for all $x$ in that interval. As demonstrated in Step 3, $f'(x) > 0$ exclusively within the interval $(0, 2)$.

Final Solution: The function $y = x^2 e^{-x}$ is increasing in the interval $(0, 2)$.

Step 1: Establishing the Function and Derivative

Given the function $f(x) = \sin x$ defined on the interval $I = (0, \frac{\pi}{2})$.

To determine the monotonicity of the function, we calculate the first derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(\sin x)$

$f'(x) = \cos x$ [Per the standard derivative formula for trigonometric functions]

Step 2: Analyzing the Sign of the Derivative

For a function to be strictly increasing on an interval $(a, b)$, the condition $f'(x) > 0$ must hold for all $x \in (a, b)$ [Per the First Derivative Test for Monotonicity].

We analyze the behavior of $f'(x) = \cos x$ within the specified interval $x \in (0, \frac{\pi}{2})$:

• At $x = 0$, $\cos(0) = 1$.
• At $x = \frac{\pi}{2}$, $\cos(\frac{\pi}{2}) = 0$.
• For all values in the open interval $0 < x < \frac{\pi}{2}$, the angle $x$ lies in the First Quadrant of the Cartesian plane.

In the first quadrant, the cosine function is strictly positive, implying $\cos x > 0$ for all $x \in (0, \frac{\pi}{2})$.

Step 3: Visual Representation of the Gradient

Step 4: Logical Conclusion

Since $f'(x) > 0$ for all $x \in (0, \frac{\pi}{2})$, it follows that the function $f(x) = \sin x$ satisfies the criteria for a strictly increasing function on the given interval.

Final Solution: Because the derivative $f'(x) = \cos x$ is strictly positive for all $x$ in the interval $(0, \frac{\pi}{2})$, the function $f(x) = \sin x$ is strictly increasing in this interval.

Step 1: Establishing the Derivative

To determine the intervals of decrease, we must analyze the first derivative of the function $f(x) = 2x^3 - 3x^2 - 36x + 7$. A function $f(x)$ is strictly decreasing on an interval when $f'(x) < 0$ [Per the Monotonicity Test].

Differentiating with respect to $x$ using the Power Rule $\frac{d}{dx}(x^n) = nx^{n-1}$:

$f'(x) = \frac{d}{dx}(2x^3) - \frac{d}{dx}(3x^2) - \frac{d}{dx}(36x) + \frac{d}{dx}(7)$

$f'(x) = 6x^2 - 6x - 36$

Step 2: Identifying Critical Points

Critical points occur where $f'(x) = 0$. We set the derivative to zero and solve for $x$:

$6x^2 - 6x - 36 = 0$

Divide the entire equation by 6:

$x^2 - x - 6 = 0$

Factor the quadratic expression: $(x - 3)(x + 2) = 0$ [By the Zero Product Property].

The critical points are $x = 3$ and $x = -2$.

Step 3: Interval Analysis

The critical points divide the real number line into three distinct disjoint intervals: $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$. We test the sign of $f'(x) = 6(x-3)(x+2)$ in each interval:

Step 4: Geometric Interpretation

Based on the analysis, $f'(x) < 0$ strictly within the interval $(-2, 3)$.

Final Solution: The function $f(x)$ is decreasing in the interval $(-2, 3)$.

Step 1: Differentiation of the Given Function

Given the function $f(x) = x^{100} + \sin x - 1$, we determine the intervals of decrease by finding the sign of its first derivative, $f'(x)$.

Applying the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the trigonometric derivative $\frac{d}{dx}(\sin x) = \cos x$ [By the Linearity of Derivatives]:

$f'(x) = \frac{d}{dx}(x^{100}) + \frac{d}{dx}(\sin x) - \frac{d}{dx}(1)$

$f'(x) = 100x^{99} + \cos x$

Step 2: Analysis of the Sign of $f'(x)$

A function $f(x)$ is strictly decreasing on an interval $I$ if $f'(x) < 0$ for all $x \in I$ [Per the First Derivative Test for Monotonicity].

We analyze the behavior of $f'(x) = 100x^{99} + \cos x$ within the interval $(0, 1)$:

• For $x \in (0, 1)$, $x^{99}$ is positive. Therefore, $100x^{99} > 0$.
• For $x \in (0, 1)$, $\cos x$ is also positive (since $(0, 1)$ radians is in the first quadrant where all trigonometric functions are positive).
• Summing two positive values: $100x^{99} + \cos x > 0$. Thus, the function is strictly increasing on $(0, 1)$.

Step 3: Evaluation of the Interval $(0, 1)$ vs $(\pi/2, \pi)$

Consider the interval $(\pi/2, \pi) \approx (1.57, 3.14)$:

• In this interval, $x > 1$. Consequently, $100x^{99}$ is a very large positive number (exceeding 100).
• The minimum value of $\cos x$ in this interval is approximately $-1$ (as $x \to \pi$).
• Since $100x^{99} > 100$ and $\cos x \geq -1$, the sum $100x^{99} + \cos x$ remains positive.

Step 4: Logical Conclusion on the Interval of Decrease

For $f(x)$ to be decreasing, we require $100x^{99} < -\cos x$. Given that the magnitude of $100x^{99}$ grows exponentially for $x > 1$ and remains significant even for small positive $x$, there is no interval of the form $(a, b)$ within the positive domain where $f'(x)$ is consistently negative.

If the question implies standard multiple-choice options provided in the CBSE curriculum (typically: $(0, 1)$, $(\pi/2, \pi)$, $(0, \pi/2)$, or None of these), we conclude through the rigorous check above that the derivative is positive for all $x > 0$.

Final Solution: The function $f(x) = x^{100} + \sin x - 1$ is strictly increasing for all $x > 0$; therefore, it is not decreasing on any of the standard intervals provided in the context of this problem, leading to the conclusion: None of these.

Step 1: Establishing the Function and Derivative

To determine the monotonicity of the function $f(x) = e^{2x}$ on the set of real numbers $\mathbb{R}$, we must examine the sign of its first derivative $f'(x)$.

The given function is: $$f(x) = e^{2x}$$ [By the Chain Rule of differentiation, where $\frac{d}{dx}[e^{u(x)}] = e^{u(x)} \cdot u'(x)$]: $$f'(x) = \frac{d}{dx}(e^{2x}) = e^{2x} \cdot \frac{d}{dx}(2x)$$ $$f'(x) = e^{2x} \cdot 2 = 2e^{2x}$$

Step 2: Analysis of the Derivative

We analyze the behavior of $f'(x) = 2e^{2x}$ for all $x \in \mathbb{R}$.

• The exponential function $e^u$ is strictly positive for all real values of $u$ ($e^u > 0, \forall u \in \mathbb{R}$).
• Since $x \in \mathbb{R}$, it follows that $2x \in \mathbb{R}$.
• Therefore, $e^{2x} > 0$ for all $x \in \mathbb{R}$.
• Multiplying by the constant $2$ (which is positive) preserves the inequality: $2e^{2x} > 0$.

Thus, $f'(x) > 0$ for all $x \in \mathbb{R}$.

Step 3: Visual Representation of $f(x) = e^{2x}$

Step 4: Theoretical Conclusion

According to the First Derivative Test for Monotonicity:

• If $f'(x) > 0$ for all $x$ in an interval $I$, then the function $f$ is strictly increasing on that interval. [Definition of Strictly Increasing Functions].

Final Solution: Since $f'(x) = 2e^{2x} > 0$ for all $x \in \mathbb{R}$, the function $f(x) = e^{2x}$ is strictly increasing on $\mathbb{R}$.

Given Function

Let the function be denoted by $f(x)$:

$f(x) = 6 - 9x - x^2$

Step 1: Differentiation of the Function

To determine the intervals of monotonicity, we compute the first derivative with respect to $x$ [Per the Power Rule of differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$]:

$f'(x) = \frac{d}{dx}(6) - \frac{d}{dx}(9x) - \frac{d}{dx}(x^2)$

$f'(x) = 0 - 9 - 2x$

$f'(x) = -(9 + 2x)$

Step 2: Finding Critical Points

The critical points occur where $f'(x) = 0$ [By the First Derivative Test criteria for critical values]:

$-(9 + 2x) = 0$

$9 + 2x = 0$

$2x = -9$

$x = -4.5$

Step 3: Interval Analysis

The critical point $x = -4.5$ divides the real number line $\mathbb{R}$ into two disjoint intervals: $(-\infty, -4.5)$ and $(-4.5, \infty)$. We analyze the sign of $f'(x) = -(9 + 2x)$ in these intervals:

Geometric Interpretation

Justification

By the Monotonicity Theorem:

• A function is strictly increasing on an interval $I$ if $f'(x) > 0$ for all $x \in I$.
• A function is strictly decreasing on an interval $I$ if $f'(x) < 0$ for all $x \in I$.

Final Solution: The function is strictly increasing on $(-\infty, -4.5)$ and strictly decreasing on $(-4.5, \infty)$.