absolute valuecalculusextreme-value-theorem
(Q) Find the extreme values for $f(x)=|3x-5|$ on $-3≤x≤2$
Because this is an absolute function $f'(x)=0$ does not exist.
There is a local minimum at $x=\frac{5}{3}$
For the interval $-3≤x≤2$, $$f(-3)=14$$ $$f(2)=1$$
Thus, the extreme values are local maximum at $(-3,14)$ and local minimum at $(\frac{5}{3}, 0)$
Is this correct?
$f'(t) = \sqrt{9-t^2} - \dfrac{t^2}{\sqrt{9-t^2}} = \dfrac{9-2t^2}{\sqrt{9-t^2}} = 0$ if $ 9-2t^2=0$ or $t = \pm\dfrac{3}{\sqrt{2}}$. But only $t = \dfrac{3}{\sqrt{2}}$ can be accepted since $-\dfrac{3}{\sqrt{2}} < -1$. Thus:
$f_{max} = \max\{f(-1), f(\frac{3}{\sqrt{2}}), f(3)\} = \max\{-2\sqrt{2},0, \frac{9}{2}\} = \dfrac{9}{2}$, and this maximum value occurs at $t = \dfrac{3}{\sqrt{2}}$, and $f_{min} = -2\sqrt{2}$ which occurs when $t = -1$.
Hint: Split up $(-1,3)$ into two intervals: $(-1,0)$ and $(0,3)$. On the former interval, since $t<0$, $|t|=-t$ and thus $f(t) = 2 + t$ on that interval. Similarly, you can show $f(t) = 2 - t$ on the second interval.
Then consider the derivative of each function on each interval, and particularly what that derivative means in the scope of the function overall - derivative tests aren't going to help a whole lot here, so think moreso about what the derivative means in a qualitative sense and how it describes the behavior of the function.
Looking at a graph might prove useful for this qualitative analysis in particular so here's one I quickly hashed up on Desmos:
Best Answer
Essentially, yes. Since you have a function of the form $f(x) = |ax+b|$, you know $f(x) \ge 0$ for all $x$. Thus, the minimum would occur wherever $f(x) = 0$, on the premise it's in the interval -- and checking it, you indeed determine that is the case. Moreover, you know that minimum will be unique, and thus any maxima will occur at other $x$ values. Since $f'(x) = \pm a$ (the sign depending on which side of the minimum you're on), you can also immediately see that such an $f$ will have maxima at the endpoints of the interval, and thus checking them is sufficient to find them: just take the greater of the two.
You can also verify your answer by graphing your function $f$ in Desmos or other software, if you please: