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Q1(iii):
Find the maximum and minimum values, if any, of the following functions given by (iii) $f(x) = – (x – 1)^2 + 10$

Solution :

Step 1: Defining the Function and Domain

The given function is $f(x) = -(x - 1)^2 + 10$. Since no specific interval $[a, b]$ is provided for the domain, we assume the domain $D$ to be the set of all real numbers, $x \in \mathbb{R}$.

Step 2: Analytical Derivation of Extrema

We analyze the behavior of the function by examining its components. The term $(x - 1)^2$ is a squared real expression. By the property of real numbers, the square of any real number is non-negative [Axiom of Real Numbers]:

$(x - 1)^2 \geq 0$ for all $x \in \mathbb{R}$

Multiplying by $-1$ reverses the inequality sign [Property of Inequalities]:

$-(x - 1)^2 \leq 0$

Adding $10$ to both sides maintains the inequality:

$-(x - 1)^2 + 10 \leq 0 + 10$

$f(x) \leq 10$

Step 3: Identification of Maximum Value

The inequality $f(x) \leq 10$ indicates that the function never exceeds $10$. The value $10$ is achieved when the term $-(x - 1)^2$ equals $0$, which occurs at $x = 1$.

Therefore, $f(1) = -(1 - 1)^2 + 10 = 10$.

By definition, since $f(x) \leq f(1)$ for all $x \in \mathbb{R}$, $f(x)$ has a global maximum of $10$ at $x = 1$.

Step 4: Examination of Minimum Value

As $x \to \infty$, $-(x - 1)^2 \to -\infty$. Consequently, $f(x) \to -\infty$. As $x \to -\infty$, $-(x - 1)^2 \to -\infty$. Consequently, $f(x) \to -\infty$.

Since the function decreases without bound as $x$ increases or decreases, there exists no real number that serves as a global minimum. Thus, the function has no minimum value.

Step 5: Graphical Representation

x y (1, 10)

Final Solution: The function has a maximum value of $10$ at $x = 1$, and it has no minimum value.


More Questions from Class 12 Mathematics Application of Derivatives EXERCISE 6.3


CBSE Solutions for Class 12 Mathematics Application of Derivatives


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