The function f(x) = |x| + |x - 1| is | vedclass

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

Question

(A) We have $f(x) = |x| + |x - 1|$.By definition of absolute value,we can write $f(x)$ as:$f(x) = \begin{cases} -2x + 1, & x < 0 \ 1, & 0 \le x < 1 \

Vedclass · Accepted Answer

Continuous at $x = 1$,but not differentiable at $x = 1$

Vedclass · Answer

(A) We have $f(x) = |x| + |x - 1|$.By definition of absolute value,we can write $f(x)$ as:$f(x) = \begin{cases} -2x + 1, & x For continuity at $x = 1$:$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} (1) = 1$$\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^+} (2x - 1) = 2(1) - 1 = 1$$f(1) = 2(1) - 1 = 1$Since $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$,the function is continuous at $x = 1$.For differentiability at $x = 1$:Left-hand derivative $(LHD)$ at $x = 1$ is $\lim_{h 	o 0^-} \frac{f(1+h) - f(1)}{h} = \frac{d}{dx}(1) = 0$.Right-hand derivative $(RHD)$ at $x = 1$ is $\lim_{h 	o 0^+} \frac{f(1+h) - f(1)}{h} = \frac{d}{dx}(2x - 1) = 2$.Since $LHD 
eq RHD$,the function is not differentiable at $x = 1$.

The function $f(x) = |x| + |x - 1|$ is

Similar Questions

Let $f(x) = |x-3| + |x+5|$ and $A = \{a \in \mathbb{R} \mid \lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a} \text{ exists} \}$. Then the number of real numbers which are in $(-\infty, -3) \cup (5, \infty)$ but not in $A$ is

If $f(x)=\frac{2x}{4+3|x|}, x \in R$, then $f^{\prime}(0)=$

The derivative of $f(x) = |x|^3$ at $x = 0$ is

Let $\alpha, \beta \in R$ be such that the function $f(x) = \begin{cases} 2 \alpha (x^2 - 2) + 2 \beta x, & x < 1 \\ (\alpha + 3) x + (\alpha - \beta), & x \ge 1 \end{cases}$ is differentiable at all $x \in R$. Then $34(\alpha + \beta)$ is equal to

Which of the following functions is differentiable at $x = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module