Prove a function is not differentiable using continuity 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
 A: $$
f(x) = |(2x)^3 - 1^3| = |2x - 1|\times |4x^2 + 2x + 1| =\\= 2|x - 1/2|\times|(1+x)^2 + 3x^2| = 2|x-1/2|(3x^2 + (1+x)^2)
$$
Since $f(1/2)=0$, you have
$$
\frac{f(x)-f(1/2)}{x-1/2} = 2\frac{|x-1/2|}{x-1/2}(3x^2 + (1+x)^2)
$$
If $x=1/2$ then $2(3x^2 + (1+x)^2) = 6$. I think now you understand why function isn't differentiable ($|t|/t=\pm1$).
A: I prefer to rewrite this as
$$g(h) =\frac{f(\frac{1}{2} +h) - f(\frac{1}{2})}{h}$$
Now, 
$$g(h) = \text{sign}(h) \left| \frac{8(\frac{1}{2} +h)^3-1}{h}  \right|$$
$$ = \text{sign}(h) \left| \frac{1 + 6h + 12h^2 +8h^3-1}{h}  \right|$$
$$ = \text{sign}(h) \left| 6 + 12h +8h^2  \right|$$
Now you can make $h\to0^+$ and  $h\to0^-$ 
A: Absolute value expressions of the form $\left|BLAH\right|$ can be simplified: when the inside is positive it equals $BLAH$, and when the inside is negative and it equals $-BLAH$.
To do the second limit, as $x \to \frac{1}{2}$ from the right side of $\frac{1}{2}$, the value of $8x^3-1$ is positive. Therefore, as $x \to \frac{1}{2}$ from the right side of $\frac{1}{2}$, using the definition of absolute value you have an equation $f(x)=8x^3-1$. So as $x \to \frac{1}{2}$ from the right side of $\frac{1}{2}$, you can evaluate the limit as 
$$\lim_{x \to \frac{1}{2}+} \frac{f(x) - f(1/2)}{x-1/2}$$
$$ = \lim_{x \to \frac{1}{2}+} \frac{(8x^3-1) - 0}{x - 1/2}$$
$$= \lim_{x \to \frac{1}{2}+} \frac{8x^3-1}{x - 1/2}
$$ 
which I'm sure you can calculate to equal $6$.
The one where $x \to \frac{1}{2}$ from the left side of $\frac{1}{2}$ is similar, the only difference being that the value of $8x^3-1$ is negative and so $f(x) = -(8x^3-1)$.
A: To see where the six comes from, we realize that if $x<1/2$ then $f(x)=-(8x^3-1)$and $\frac{-(8x^3-1)}{x-1/2}=-(8x^2+4x+2)$. Plugging in 1/2 here yields -6.
A: In calculating right limit that is when $x\to \frac{1}{2}^+$ we have $x>\frac{1}{2}$ so $(8x^3-1)>0$ therefore $\mid 8x^3-1 \mid =8x^3-1$and then in $$\lim_{x \to \frac12+} \dfrac{f(x)-f(1/2)}{x-1/2} $$ both numerator  and Denominator are positive then the answer is $6.$ Since right and left derivatives are not equal then the function is not differentiable.
