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+5 votes

Consider the following two statements about the function $f(x)=\left\vert x\right\vert$:

  • P. $f(x)$ is continuous for all real values of $x$.
  • Q. $f(x)$ is differentiable for all real values of $x$ .

Which of the following is TRUE?

  1. P is true and Q is false.
  2. P is false and Q is true.
  3. Both P and Q are true.
  4. Both P and Q are false.
asked in Calculus by Veteran (59.4k points) | 991 views

2 Answers

+13 votes
Best answer

ans is A. f(x)=|x| here for all values of x, f(x) exists. therefore it is continuous for all real values of x. 

At x=0, f(x) is not differentiable. Because if we take the left hand limit here, it is negative while the right hand limit is positive.   


answered by Loyal (8.3k points)
selected by
at x= 0, Left hand derivative is not equal to right hand derivative. So it is not differentiable.

at x=0 Left hand limit and Right hand limit are equal , for derivability we check LHD(left hand derivative) and RHD(Right hand derivative) , those are not equal so function at x=0 is not differentiable.

$LHD $ = $f'(a^{-})$ = $\frac{f(a+h)-f(a)}{h}$   where $h->0^{-}$

$RHD$ = $f'(a^{+})$ = $\frac{f(a+h)-f(a)}{h}$   where $h->0^{+}$

LHD = $f'(0^{-})$ = $\frac {f(0+h)-f(0)}{h}$ = $\frac {\left | 0+h \right | -\left | 0 \right |}{h}$ = $\lim_{h->0^{-}} \frac{\left | h \right |}{h} = \frac{-h}{h}=-1$

RHD = $f'(0^{+})$ = $\frac {f(0+h)-f(0)}{h}$=$\frac {\left | 0+h \right | -\left | 0 \right |}{h}$ = $\lim_{h->0^{+}} \frac{\left | h \right |}{h} = \frac{h}{h}=1$

$LHD\neq RHD$

0 votes
A) is true . This function is not diffrentiable at x=0
answered by (57 points)

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