Consider a function $$f(x)=\cos(|x-5|)-\sin(|x-3|)+|x+10|^3-(|x|+4)^2.$$
At which points f is not differentiable? a. $x=5$, b. $x=3$, c. $x=-10$, d. $x=0$.

I know that $|x|$ is not differentiable at $x=0$ hence d. is correct but how to check differentiability of other function?

  • $\begingroup$ Write down the function when $3 < x < 5$. This is $f(x) = \cos(5-x) - \sin(x-3) + (x+10)^3 -(x+4)^2$ and when $x > 5$, the function is $f(x) = \cos(x-5) - \sin(x-3) + (x+10)^3 -(x+4)^2$ Since the derivative of these two expressions evaluate to the same value at $x=5$, the function is differentiable at $x=5$ $\endgroup$ – user348749 Dec 7 '16 at 7:04

Hint. Show that:

i) $\displaystyle \lim_{x\to 5}\frac{\cos(|x-5|)-1}{x-5}=0$,

ii) $\displaystyle \lim_{x\to 3^{\pm}}\frac{\sin(|x-3|)}{x-3}=\pm1$,

iii) $\displaystyle \lim_{x\to -10^{\pm}}\frac{|x+10|^3}{x+10}=\pm1$.

Can you take it from here?

  • $\begingroup$ I get it. Thanks. $\endgroup$ – ASHWINI SANKHE Dec 7 '16 at 8:09

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