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I am a little confused on how to find the absolute max and min without using a calculator. I know I have to find the derivative of this function and find critical values, but I'm not finding any critical values. I know that the max value of this function in the given interval is $(-5,5)$, but I cannot find the minimum. Any suggestions?

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    $\begingroup$ The function $f(x)=|x|$ looks like a $v$, now it is translated to the right by $3$ and down with $3$. The maximum value is therefor in the leftmost point, i.e. $|-5-3|-3 = 5$ and the minimum is in $x=3$ namely $-3$. $\endgroup$ – Math Dec 15 '19 at 18:23
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The critical values are the x values where $n'$ is either $0$ or undefined. There is no point at which $n'(x)=0$, but $n'(3)$ is undefined. Therefore, the maximum and minimum values can only occur when $x\in\{-5,3,5\}$. Since $n(-5)=5$, $n(3)=-3$, and $n(5)=-1$, the absolute maximum is $5$ and the absolute minimum is $-3$.

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  • $\begingroup$ I see, I totally forgot about when the derivative is undefined. Thank You! $\endgroup$ – P1081 Dec 15 '19 at 18:25
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Start by considering the absolute minimum of

$f(x)= |x|$.

The minimum is $0$ at $x = 0$. Then, look at

$f(x) = |x - 3|$.

The minimum is still $0$, now at $x = 3$. Lastly, the minimum of $n(x)$ is $-3$ at $n(3)$.

Unless your teacher tells you differently, it’s best to answer questions asking for minimums or maximum values by actually writing the value and where it occurs, rather than giving an ordered pair.

Instead of writing $n(x)$’s absolute maximum is at $(-5,5)$ as you did, write “$n(x)$ has an absolute maximum of $5$ at $x=-5$.” You could also write “the function has an absolute maximum of $5$ at $n(-5)$.” Note that inflection points are a related concept, but are actually points so when you give an inflection point you will give an ordered pair.

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The absolute minimum is $-3$, as it can be reached by $n$ ($n(3)=-3$), and also $\forall x \in [-5,5]:n(x)=|x-3|-3\geq -3$.

The absolute maximum is $5$, as it can by reached by $n$ ($n(-5)=5$). Also $\forall x\in [-5,3]:n(x)=-x+3-3=-x\leq 5$ (since $-x$ is decreasing and plugging in $-5$ gives $5$) and $\forall x \in [3,5]: n(x)=x-3-3=x-6\leq -1$ (since $x-6$ is increasing and plugging in 5 gives $-1$).

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With differentiable functions, minima occur at either the boundaries or where the derivative changes from negative to positive. The absolute value function is not differentiable, but you will still find that the minima occurs at a point where the derivative changes from negative to positive.

BTW, the term "critical point" is sometimes defined as "points where the derivative is zero", but a better definition is "points where the derivative doesn't have a value other than zero". That definition includes points where the derivative doesn't have any value, because if it doesn't have a value, it doesn't have a value other than zero. Points where the derivative is undefined, or you can only take a one-sided derivative because it's a boundary point, should be included. The reason this works for finding minima is that if the point has a positive derivative, then the function is decreasing as you decrease $x$, so there are smaller values to the left, so it's not a minimum. If the derivative is negative, then there are smaller values to the right. The only way there can't be smaller values is if the derivative is neither positive nor negative, which leaves zero or undefined.

However, finding this point isn't significantly less difficult (and arguably is more difficult) than simply finding the minimum directly. This is a situation where Calculus doesn't really add much to the tools from Alg II. If you can graph the function (and if you passed Alg II, you should be able to), then you should be able to answer this question. Just because you just learned how to use hammers, doesn't mean you should assume everything you come across is a nail.

Simply ask yourself: what are the special points? By inspection, the special points are $x = -5$ (left side boundary), $x = 5$ (right side boundary), and $x = 3$ (the stuff inside the absolute value function changes signs). So those are the three candidates for maximum and minimum.

When you take the absolute value, you never get a negative result, so the least it can be is zero. So the minimum of $|x-3|$ is at least $0$, and the minimum of $|x-3|-3$ is at least $-3$. You then have to check whether it reaches that value. For what value of $x$ is $|x-3|$ equal to zero, and is that $x$ in the interval? If that value of $x$ isn't in the interval, it follows that the minimum occurs at a boundary point.

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