Let f(x) = begin{cases} |x|, & -infty

Question

(B) We define the function $f(x)$ by analyzing the absolute values: $|x| = \begin{cases} -x, & x < 0 \ x, & x \geq 0 \end{cases}$ and $|2x-4| = \begi

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) We define the function $f(x)$ by analyzing the absolute values: $|x| = \begin{cases} -x, & x Thus,$f(x) = \begin{cases} -x, & -\infty At $x=0$: $	ext{LHL} = \lim_{x ightarrow 0^-} (-x) = 0$,$	ext{RHL} = \lim_{x ightarrow 0^+} (x) = 0$,and $f(0) = 0$. Since $	ext{LHL} = 	ext{RHL} = f(0)$,$f(x)$ is continuous at $x=0$. However,the left derivative is $-1$ and the right derivative is $1$,so it is not differentiable at $x=0$. At $x=2$: $	ext{LHL} = \lim_{x ightarrow 2^-} (x) = 2$,$	ext{RHL} = \lim_{x ightarrow 2^+} (2x-4) = 0$. Since $	ext{LHL} 
eq 	ext{RHL}$,$f(x)$ is not continuous at $x=2$,and therefore not differentiable at $x=2$. Thus,$a=0$ (continuous but not differentiable) and $b=2$ (not differentiable). Therefore,$a+b = 0+2 = 2$.

Let $f(x) = \begin{cases} |x|, & -\infty < x < 2 \\ |2x-4|, & 2 \leq x \leq 20 \end{cases}$. If $x=a$ is a point where $f(x)$ is continuous but not differentiable and $x=b$ is a point where $f(x)$ is not differentiable $(a \neq b)$,then $a+b=$

Similar Questions

If $f(x) = \begin{cases} x \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$,then at $x = 0$ the function $f(x)$ is

$f(x)= \begin{cases}(1+3x)^{\frac{4}{x}}, & \text{if } x \neq 0 \\ a, & \text{if } x=0 \end{cases}$
If $f$ is continuous at $x=0$,then $\log a=$

The function $f(x) = 2x - |x - x^2|$ is

Let $f(x)=\lim _{\theta \rightarrow 0}\left(\frac{\cos \pi x-x^{\left(\frac{2}{\theta}\right)} \sin (x-1)}{1+x^{\left(\frac{2}{\theta}\right)}(x-1)}\right), x \in R$. Consider the following two statements: $(I)$ $f(x)$ is discontinuous at $x=1$. $(II)$ $f(x)$ is continuous at $x=-1$. Then,

Let $f: R \rightarrow R$ be a continuous function such that $f(x^2) = f(x^3)$ for all $x \in R$. Consider the following statements:
$I.$ $f$ is an odd function.
$II.$ $f$ is an even function.
$III.$ $f$ is differentiable everywhere.
Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module