Let f(x) = begin{cases} |x| & text{for } 0

Question

(A) Given the function $f(x) = \begin{cases} |x| & 0  0$. Sinc

Vedclass · Accepted Answer

Local maximum.

Vedclass · Answer

(A) Given the function $f(x) = \begin{cases} |x| & 0 For $x$ close to $0$ (but $x 
eq 0$),$|x| > 0$. Since $f(x) = |x|$ for $x 
eq 0$,we have $f(x) > 0$ for all $x$ in the neighborhood of $0$ except at $x=0$.At $x = 0$,$f(0) = 1$.For any $x$ in the interval $(-\delta, \delta)$ where $\delta$ is small (e.g.,$\delta = 0.5$),we have $f(x) = |x|$.Since $|x| Specifically,for $x \in (-0.5, 0.5)$ and $x 
eq 0$,$f(x) = |x| Since $f(x) < f(0)$ for all $x$ in the neighborhood of $0$,$x = 0$ is a point of local maximum.

Let $f(x) = \begin{cases} |x| & \text{for } 0 < |x| \leqslant 2 \\ 1 & \text{for } x = 0 \end{cases}$. What is the nature of $f$ at $x = 0$?

Similar Questions

On the interval $[0, 1],$ the function $f(x) = {x^{25}}{(1 - x)^{75}}$ takes its maximum value at the point

Let the sum of the maximum and the minimum values of the function $f(x) = \frac{2x^2 - 3x + 8}{2x^2 + 3x + 8}$ be $\frac{m}{n}$,where $\gcd(m, n) = 1$. Then $m + n$ is equal to :

The maximum area of a rectangle that can be inscribed in a circle of radius $2 \text{ units}$ is

The function $f(x) = \sin^p x \cos^q x$ has a maximum at:

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan ^{-1} \sqrt{2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module