Proving a function is not differentiable
Given the function $f(x) = |8x^3 − 1|$ in the set $A = [0, 1].$ Prove that the function is not differentiable at $x = \frac12.$
The answer in my book is as follows:
$$\lim_{x \to \frac12-} \dfrac{f(x)-f(1/2)}{x-1/2} = -6$$ $$\lim_{x \to \frac12+} \dfrac{f(x)-f(1/2)}{x-1/2} = 6$$
Can anyone explain how the $6$'s were derived. I understand that as $x$ tends to $\frac12$ from the negative side, the bottom will be negative, so thats why the first one is a minus.
But how do you get to the $6$, what am I missing? Obviously $f(\frac12)=0$ but what do you make $f(x)=$ as $x$ tends to $\frac12$
Thanks
$\endgroup$4 Answers
$\begingroup$Let $f_+(x)= 8x^3-1$, $f_-(x) = 1-8x^3$, these are both smooth.
In the following $f$ is the function the OP defined. (I am not redefining $f$.)
If $x \ge { 1\over 2}$ then $f(x) = f_+(x)$.
If $x \le { 1\over 2}$ then $f(x) = f_-(x)$.
Hence the one sided derivatives of $f$ will match the derivatives of $f_+,f_-$ as appropriate.
$\endgroup$ 8 $\begingroup$Hint: $f(x) = |g(x)|$ where $g(x) = 8x^3 - 1$. What is the derivative of $g(x)$ when $x = \frac{1}{2}$?
$\endgroup$ $\begingroup$$(8x^3-1)/(x-1/2)=8(x^3-1/8)/(x-1/2)=8(x^2+(1/2)x+1/4)$, which is equal to 6 at $x=1/2$.
$\endgroup$ $\begingroup$It's almost easier to prove a more general result: If $f(a) = 0$ and $f'(a) \ne 0,$ then $|f(x)|$ is not differentiable at $a.$
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