Identify the correct statement about the func | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To analyze $f(x) = \max(x^2 - 1, 7 - x^2, 5)$,we find the intersection points of the curves:$1$. $x^2 - 1 = 7 - x^2 \implies 2x^2 = 8 \implies x^2

Vedclass · Accepted Answer

$f(x)$ is not differentiable at $4$ points.

Vedclass · Answer

(A) To analyze $f(x) = \max(x^2 - 1, 7 - x^2, 5)$,we find the intersection points of the curves:$1$. $x^2 - 1 = 7 - x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2$. At $x = \pm 2$,$f(x) = 3$.$2$. $x^2 - 1 = 5 \implies x^2 = 6 \implies x = \pm \sqrt{6}$.$3$. $7 - x^2 = 5 \implies x^2 = 2 \implies x = \pm \sqrt{2}$.Comparing the values,the function $f(x)$ is defined as:$f(x) = 7 - x^2$ for $x \in [-\sqrt{2}, \sqrt{2}]$$f(x) = 5$ for $x \in [-\sqrt{6}, -\sqrt{2}] \cup [\sqrt{2}, \sqrt{6}]$$f(x) = x^2 - 1$ for $x \in (-\infty, -\sqrt{6}] \cup [\sqrt{6}, \infty)$The function is continuous everywhere. The non-differentiable points occur at the intersection points $x = \pm \sqrt{2}$ and $x = \pm \sqrt{6}$,which are $4$ points.Thus,$f(x)$ is not differentiable at $4$ points. The correct option is $A$.

Identify the correct statement about the function $f(x) = \max(x^2 - 1, 7 - x^2, 5)$.

Similar Questions

Let $f(x) = \begin{cases} \sin x, & \text{for } x \ge 0 \\ 1 - \cos x, & \text{for } x \le 0 \end{cases}$ and $g(x) = e^x$. Then $(g \circ f)'(0)$ is

If $f(x) = \begin{cases} 2a - x & \text{when } -a < x < a \\ 3x - 2a & \text{when } a \leq x \end{cases}$,then which of the following is true?

Let $f(x) = \frac{x}{1 + |x|}$ be differentiable at . . . .

The left-hand derivative of $f(x) = [x]\sin(\pi x)$ at $x = k$,where $k$ is an integer and $[x]$ denotes the greatest integer function $\le x$,is:

Let a function $g:[0,4] \rightarrow R$ be defined as
$g(x) = \begin{cases} \max_{0 \leq t \leq x} \{t^3 - 6t^2 + 9t - 3\} & , 0 \leq x \leq 3 \\ 4 - x & , 3 < x \leq 4 \end{cases}$
Then the number of points in the interval $(0,4)$ where $g(x)$ is $NOT$ differentiable is $.....$

Vedclass Test Series

Exam Paper Generator

Online Exam Module