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Show that the function \( f(x) = |x+2 | \) is continuous at every \( x \in R\) but fails to be differentiable at x = -2.

1 Answer

  • If $f$ is a real function on a subset of the real numbers and $c$ a point in the domain of $f$, then $f$ is continous at $c$ if $\lim\limits_{x\to c} f(x) = f(c)$.
  • Every polynomial function $f(x)$ is continuous.
Step 1:
$f(x)=\mid x+2\mid$
$f(x)=\left\{\begin{array}{1 1}-(x+2),&for\;x< -2\\(x+2)&for\;x\geq-2\end{array}\right.$
Let us check the differentiability at $x=-2$
LHD at $x=1=\lim\limits_{x\to 2^-}\large\frac{f(x)-f(-2)}{x+2}$
$\Rightarrow \lim\limits_{x\to 2^-}\large\frac{-(x+2)-0}{x+2}$
$\Rightarrow \lim\limits_{x\to 2^-}\large\frac{-x-2}{x+2}=\large\frac{-(x+2)}{x+2}$
$\Rightarrow -1$
Step 2:
RHD at $x=-2$ is
$\lim\limits_{x\to 2^+}\large\frac{f(x)-f(-2)}{x+2}$
$\Rightarrow 1$
Hence LHD $\neq$ RHD
Hence f(x) is not differentiable at $x=-2$
answered Nov 8, 2013 by sreemathi.v

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