Let f(x)=15-|x-10| ; x in R. Then,the set of | vedclass

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

Question

(D) Given $f(x) = 15 - |x - 10|$.We need to find the points where $g(x) = f(f(x))$ is not differentiable.First,$f(x)$ is not differentiable at $x = 10

Vedclass · Accepted Answer

$\{5,10,15\}$

Vedclass · Answer

(D) Given $f(x) = 15 - |x - 10|$.We need to find the points where $g(x) = f(f(x))$ is not differentiable.First,$f(x)$ is not differentiable at $x = 10$.Now,$g(x) = f(f(x)) = 15 - |f(x) - 10| = 15 - |(15 - |x - 10|) - 10| = 15 - |5 - |x - 10||$.The function $g(x)$ is not differentiable where the inner expressions are zero or where the original function $f(x)$ is not differentiable.$1$. $f(x)$ is not differentiable at $x = 10$.$2$. The expression inside the outer modulus,$5 - |x - 10|$,is zero when $|x - 10| = 5$,which gives $x - 10 = 5$ or $x - 10 = -5$,so $x = 15$ or $x = 5$.Thus,the function $g(x)$ is not differentiable at $x \in \{5, 10, 15\}$.

Let $f(x)=15-|x-10| ; x \in R$. Then,the set of all values of $x$,at which the function $g(x)=f(f(x))$ is not differentiable,is

Similar Questions

Which of the following radioisotopes is used as anticancerous?

$\cos A \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A$ equals

If the harmonic mean of the roots of the equation $\sqrt{2}x^2 - bx + (8 - 2\sqrt{5}) = 0$ is $4$,then the value of $b$ is:

Let $P = \{\theta : \sin \theta - \cos \theta = \sqrt{2} \cos \theta\}$ and $Q = \{\theta : \sin \theta + \cos \theta = \sqrt{2} \sin \theta\}$ be two sets,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module