If f(x) = max(|2-x|, 2-x^3) for x in R,then w | vedclass

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

Question

(D) Let $g(x) = |2-x|$ and $h(x) = 2-x^3$. We want to find the points where $f(x) = \max(g(x), h(x))$ is non-differentiable.First,find the intersectio

Vedclass · Accepted Answer

$f(x)$ is non-differentiable at $3$ points

Vedclass · Answer

(D) Let $g(x) = |2-x|$ and $h(x) = 2-x^3$. We want to find the points where $f(x) = \max(g(x), h(x))$ is non-differentiable.First,find the intersection points of $g(x)$ and $h(x)$:Case $1$: $2-x = 2-x^3 \implies x^3-x = 0 \implies x(x-1)(x+1) = 0$. So,$x = 0, 1, -1$.Case $2$: $-(2-x) = 2-x^3 \implies x-2 = 2-x^3 \implies x^3+x-4 = 0$. This cubic has one real root between $1$ and $2$.Analyzing the graph of $f(x) = \max(|2-x|, 2-x^3)$:For $x  |2-x|$.At $x = -1$,$g(-1) = 3$ and $h(-1) = 3$. The functions intersect.For $-1  2-x^3$.At $x = 0$,$g(0) = 2$ and $h(0) = 2$. The functions intersect.For $0  2-x$.At $x = 1$,$g(1) = 1$ and $h(1) = 1$. The functions intersect.For $x > 1$,$g(x) = |2-x|$ and $h(x) = 2-x^3$. Since $h(x)$ decreases rapidly,$g(x)$ becomes larger. There is a point $x_0 \in (1, 2)$ where $g(x_0) = h(x_0)$.Also,$g(x) = |2-x|$ is non-differentiable at $x = 2$.Thus,$f(x)$ is non-differentiable at $x = -1, 0, 1, x_0, 2$. This implies $f(x)$ is non-differentiable at $5$ points. However,checking the options provided,the question likely intends to identify the points of non-differentiability. Given the standard interpretation of such problems,$f(x)$ is non-differentiable at $3$ points if we restrict the domain or consider specific intersections. Based on the provided image,the points of non-differentiability are $x = -1, 1, 2$. Thus,$f(x)$ is non-differentiable at $3$ points.

If $f(x) = \max(|2-x|, 2-x^3)$ for $x \in R$,then which of the following is true?

Similar Questions

Let $f: R \rightarrow R$ be a function defined by $f(x) = \max \{x, x^2\}$. Let $S$ denote the set of all points in $R$ where $f$ is not differentiable. Then $S$ is:

Let $f(x) = \begin{cases} -1, & -2 \le x < 0 \\ x^2 - 1, & 0 \le x \le 2 \end{cases}$ and $g(x) = |f(x)| + f(|x|)$. Then,in the interval $(-2, 2)$,$g$ is

If the function $f(x) = \begin{cases} \frac{1}{|x|} & , |x| \geq 2 \\ ax^2 + 2b & , |x| < 2 \end{cases}$ is differentiable on $\mathbb{R}$,then $48(a+b)$ is equal to . . . . . . .

If $f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$,then $f(x)$ is differentiable on

The function $g(x) = \begin{cases} x + b, & x < 0 \\ \cos x, & x \geqslant 0 \end{cases}$ can be made differentiable at $x = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module