Please help, the absolute value throws me off.
$\displaystyle g(t) = |3t-4|$
You start with definitions! A "critical point" of a function of one variable is a point where either the derivative is 0 or the derivative does not exist.
The |x| is x if $\displaystyle x\ge 0$ and -x if $\displaystyle x< 0$.
In particular |3t- 4| is 3t- 4 if $\displaystyle 3t- 4\ge 0$ or $\displaystyle t\ge 4/3$, -(3t- 4)= 4- 3t for [tex]t< 4/3[/tex[. So for [tex]t> 4/3[tex], the function is 3t- 4 and has derivative 3. For $\displaystyle x< 3/4$, the function is 4- 3t which has derivative -3. Those are never 0 so the only possible critical point would be at x= 4/3. We need to look at that point separately.
To determine the derivative at x= 4/3 we can do either of two things:
1) Use the basic definition. The derivative of f(x) at x= 4/3 is $\displaystyle \lim_{h\to 0}\frac{f((4/3)+h)- f(4/3)}{h}$. f(4/3)= 0, of course, but because of the change of formula on both sides of x= 4/3, we need to look at the "one sided" derivatives. If h> 0 then (4/3)+ h> 4/3, f((4/3)+ h)= 3((4/3)+ h)- 4= 3h so the difference quotient is $\displaystyle \frac{3h}{h}= 3$ and the limit is 3. If h< 0 then (4/3)+ h< 4/3, f((4/3)+h)= 4- 3((4/3)+ h)= -3h so the difference quotient is $\displaystyle \frac{-3h}{h}= -3$ and the limit is -3. Since those two one sided limits are not the same, the limit itself does not exist and the function is not differentiable at x= 4/3.
2) Use a more "sophisticated" theorem. While the derivative of a differentiable function is not itself necessarily continuous, one can use the mean value theorem to show that the "intermediate value theorem" does, in fact, apply. From that, it follows that if the derivative at a specific point exists, the two limits $\displaystyle \lim_{x\to a^-}f'(x)$ and $\displaystyle \lim_{x\to a^+}f'(x)$ must both be equal to the derivative at that point. Here, of course, those two limits are -3 and 3 and are not the same so the function is not differentiable at x= 4/3.