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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 (59.5k points)
edited by | 1k views

2 Answers

+14 votes
Best answer

Ans is A.

$f(x)=\mid x\mid.$ 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 making $LHL \neq RHL$  


answered by Loyal (8.2k points)
edited 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 (115 points)

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