If f is defined by f(x) = begin{cases} x & te | vedclass

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

Question

(B) Given function,$f(x) = \begin{cases} x & 	ext{for } 0 \leq x < 1 \ 2-x & 	ext{for } x \geq 1 \end{cases}$.At $x=1$,$f(1) = 2-1 = 1$.Left Hand L

Vedclass · Accepted Answer

continuous but not differentiable

Vedclass · Answer

(B) Given function,$f(x) = \begin{cases} x & 	ext{for } 0 \leq x At $x=1$,$f(1) = 2-1 = 1$.Left Hand Limit $(LHL)$ = $\lim_{x ightarrow 1^-} f(x) = \lim_{x ightarrow 1^-} x = 1$.Right Hand Limit $(RHL)$ = $\lim_{x ightarrow 1^+} f(x) = \lim_{x ightarrow 1^+} (2-x) = 2-1 = 1$.Since $LHL$ = $RHL$ = $f(1)$,the function $f(x)$ is continuous at $x=1$.Now,to check differentiability,we find the derivative $f'(x)$:$f'(x) = \begin{cases} 1 & 	ext{for } 0 \leq x  1 \end{cases}$.Left Hand Derivative $(LHD)$ = $\lim_{x ightarrow 1^-} f'(x) = 1$.Right Hand Derivative $(RHD)$ = $\lim_{x ightarrow 1^+} f'(x) = -1$.Since $LHD$ $
eq$ $RHD$,the function is not differentiable at $x=1$.Therefore,$f(x)$ is continuous but not differentiable at $x=1$.

If $f$ is defined by $f(x) = \begin{cases} x & \text{for } 0 \leq x < 1 \\ 2-x & \text{for } x \geq 1 \end{cases}$,then at $x=1$,$f(x)$ is

Similar Questions

Which of the following statements is true for the function $f(x) = \begin{cases} \sqrt{x} & x \ge 1 \\ x^3 & 0 \le x < 1 \\ \frac{x^3}{3} - 4x & x < 0 \end{cases}$

Let $C$ be the curve $y = x^3$ (where $x$ takes all real values). The tangent at $A(t, t^3)$ meets the curve again at $B(T, T^3)$. If the gradient at $B$ is $K$ times the gradient at $A$,then $K$ is equal to

Let $a_1, a_2, \ldots, a_{100}$ be non-zero real numbers such that $a_1+a_2+\ldots+a_{100}=0$. Then,

If a function $f$ is defined by:
$\begin{cases} f(x) = x-1, & \text{when } -\infty < x < 1 \\ f(x) = 0, & \text{when } x=1 \\ f(x) = x^3-1, & \text{when } 1 < x < \infty \end{cases}$
then at $x=1$,$f$ is:

If $f(x) = \begin{cases} x^{3}-3x+2, & x < 2 \\ x^{3}-6x^{2}+9x+2, & x \geq 2 \end{cases}$, then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module