The function f(x) = (x^2 - 1)|x^2 - 3x + 2| + | vedclass

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

Question

(D) The function $f(x) = (x^2 - 1)|(x - 1)(x - 2)| + \cos(|x|)$.First,consider the term $\cos(|x|)$. The function $|x|$ is not differentiable at $x =

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) The function $f(x) = (x^2 - 1)|(x - 1)(x - 2)| + \cos(|x|)$.First,consider the term $\cos(|x|)$. The function $|x|$ is not differentiable at $x = 0$,so $\cos(|x|)$ is not differentiable at $x = 0$.Next,consider the term $(x^2 - 1)|(x - 1)(x - 2)|$. The expression $|(x - 1)(x - 2)|$ is not differentiable at the roots of $(x - 1)(x - 2) = 0$,which are $x = 1$ and $x = 2$.At $x = 1$,$f(x) = (x^2 - 1)|(x - 1)(x - 2)| + \cos(|x|)$. Since $(x^2 - 1)$ has a factor $(x - 1)$,the product $(x^2 - 1)|(x - 1)(x - 2)|$ becomes differentiable at $x = 1$ because the zero of the polynomial cancels the non-differentiability of the absolute value.At $x = 2$,the term $(x^2 - 1)$ is $3 
eq 0$. Thus,the non-differentiability of $|(x - 1)(x - 2)|$ at $x = 2$ persists.Checking $x = 2$:$Lf'(2) = \lim_{h 	o 0^-} \frac{f(2+h) - f(2)}{h} = -3 - \sin(2)$$Rf'(2) = \lim_{h 	o 0^+} \frac{f(2+h) - f(2)}{h} = 3 - \sin(2)$Since $Lf'(2) 
eq Rf'(2)$,the function is not differentiable at $x = 2$.

The function $f(x) = (x^2 - 1)|x^2 - 3x + 2| + \cos(|x|)$ is not differentiable at

Similar Questions

If $f(x) = \begin{cases} 1, & x < 0 \\ 1 + \sin x, & 0 \le x < \frac{\pi}{2} \end{cases}$,then $f'(0) = $

Let $S = \{(\lambda, \mu) \in R \times R : f(t) = (\|\lambda\|e^{\|t\|} - \mu) \sin(2\|t\|), t \in R\}$ be a differentiable function. Then $S$ is a subset of?

Assertion $(A)$: If $y = f(x) = (|x| - |x - 1|)^2$,then $\left(\frac{dy}{dx}\right)_{x=1} = 1$.
Reason $(R)$: If $\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$ exists,then it is called the derivative of $f(x)$ at $x = a$.
Then:

The function that is not differentiable at $x=1$ is

The set of points where the function $f(x)=|x-1| e^{x}$ is differentiable,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module