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Examine the differentiability of f,where f is defined by $f(x)=\left \{\begin{array}{1 1}1+x, & if\;x\leq 2\\5-x, & if\;x>2\end{array}\right.$at $x=2$.

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  • A function is not differentiable if LHL $\neq$ RHL.
  • A function is not differentiable if LHL or RHL does not exist.
Step 1:
$f(x)=\left \{\begin{array}{1 1}1+x, & if\;x\leq 2\\5-x, & if\;x>2\end{array}\right.$at $x=2$.
LHD at $x=2$
$\qquad\qquad=\lim\limits_{\large x\to 2^-}\large\frac{f(x)-f(2)}{x-2}$
$\qquad\qquad=\lim\limits_{\large x\to 2^-}\large\frac{(1+x)-(1+2)}{x-2}$
$\qquad\qquad=\lim\limits_{\large x\to 2^-}\large\frac{x-2}{x-2}$
$\qquad\qquad=1$
Step 2:
LHD at $x=2$
$\qquad\qquad=\lim\limits_{\large x\to 2^+}\large\frac{f(x)-f(2)}{x-2}$
$\qquad\qquad=\lim\limits_{\large x\to 2^+}\large\frac{(5-x)-(5-2)}{x-2}$
$\qquad\qquad=\lim\limits_{\large x\to 2^+}\large\frac{2-x}{x-2}$
$\qquad\qquad=-1$
Hence LHD at $x=2 \neq$ RHD at $x=2$.
So $f(x)$ is not differentiable at $x=2$
answered Jun 27, 2013 by sreemathi.v
 

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