Video: Find the Absolute Maximum and Minimum of a Piecewise-Defined Function

Consider the function 𝑓(π‘₯) = (π‘₯ βˆ’ 3)Β², if π‘₯ ≀ 5 and 𝑓(π‘₯) = 4π‘₯ βˆ’ 16, if π‘₯ > 5, over the interval [0, 7]. Determine the absolute minimum of 𝑓(π‘₯) over the given interval. Determine the absolute maximum of 𝑓(π‘₯) over the given interval.

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Video Transcript

Consider the function 𝑓 of π‘₯ is equal to π‘₯ minus three all squared if π‘₯ is less than or equal to five and 𝑓 of π‘₯ is equal to four π‘₯ minus 16 if π‘₯ is greater than five over the closed interval from zero to seven. Determine the absolute minimum of 𝑓 of π‘₯ over the given interval. Determine the absolute maximum of 𝑓 of π‘₯ over the given interval.

The question gives us a piecewise-defined function 𝑓 of π‘₯ defined on a closed interval. And we’re asked to find the absolute minimum and absolute maximum of our function 𝑓 of π‘₯ over this interval. We know how to find the absolute maximum and absolute minimum of a continuous function over a closed interval.

To find the absolute extrema of a continuous function on a closed interval, we follow the following three steps. First, we find all the critical points of our function. Then we evaluate our function at all of the critical points. And finally, we evaluate our function at the endpoints of our closed interval. But in this case, we’re given a piecewise function. We don’t know if this is necessarily continuous or not.

We could check if our function 𝑓 of π‘₯ is continuous and then carry on using this method, and this would work. However, there’s a more simple method for finding absolute extrema of piecewise-defined functions. We know that π‘₯ minus three all squared is continuous and four π‘₯ minus 16 is continuous. So we could find the absolute extrema on each of these pieces separately. Then we’ll just check the value of 𝑓 of π‘₯ around π‘₯ is equal to five separately. If this function has a jump discontinuity, it will have a different value as π‘₯ approaches five from the left and five from the right.

So let’s start with the first part of our piecewise-defined function 𝑓 of π‘₯. We want to find all of the extrema of π‘₯ minus three all squared on the closed interval from zero to five. And this is a continuous function on a closed interval. So we can do this by following our three steps.

First, we need to find the critical points of this function. This is where the derivative of our function is equal to zero or where the derivative does not exist. However, this is a polynomial. So its derivative exists for all real values of π‘₯. To find the derivative of π‘₯ minus three all squared, we’ll start by distributing the square over our parentheses. We get π‘₯ squared minus six π‘₯ plus nine.

Then we can differentiate this term by term by using the power rule for differentiation. We get two π‘₯ minus six. And we can then solve this equal to zero. We get that π‘₯ is equal to three. So we found one critical point of this function. It’s when π‘₯ is equal to three.

Next, we need to evaluate π‘₯ minus three all squared, that’s our critical point, and the endpoints of our interval. Evaluating this at π‘₯ is equal to three, we get three minus three all squared, which is equal to zero. Now, we evaluated our first endpoint π‘₯ is equal to zero. We get zero minus three all squared, which is equal to nine. Finally, we evaluate to the other end of our closed interval. This is π‘₯ is equal to five. So we get five minus three all squared, which is equal to four.

Let’s now find any extrema for the second part of our piecewise-defined function 𝑓 of π‘₯. We’ll look for extrema of four π‘₯ minus 16 on the closed interval from five to seven. However, it’s worth reiterating we’ve not shown that our function 𝑓 of π‘₯ is equal to four π‘₯ minus 16 when π‘₯ is equal to five. So if we get an extrema at this point, we’ll need to consider what this means graphically. But for now, we’ll just carry on and find the extrema of this continuous function by using our three steps.

First, we need to find the derivative of this linear function. Since it’s a linear function, the derivative will be the coefficient of π‘₯, which is four. And so our derivative is not equal to zero for any value of π‘₯. So we only need to evaluate this at the endpoints of our interval. This time, our closed interval is from five to seven.

We’ll start with π‘₯ is equal to five. We get four times five minus 16, which is equal to four. And it’s worth pointing out this is the same as what we got when we evaluated our other endpoint. In other words, we could’ve shown that our function 𝑓 of π‘₯ was continuous by using direct substitution on the left- and right-hand limit. But our current method will still work. We’ll now check the other endpoint of our closed interval π‘₯ is equal to seven. We get four times seven minus 16, which is equal to 12.

We’ve now evaluated our function 𝑓 of π‘₯ at all of the possible values where there could be an absolute minimum or an absolute maximum on this closed interval. We can see our lowest value was when π‘₯ is equal to three. It gave us zero. And the largest value is when π‘₯ was equal to seven. It gave us an output of 12.

So the absolute minimum of our piecewise-defined function 𝑓 of π‘₯ over the closed interval from zero to seven must be zero. And the absolute maximum of our piecewise-defined function 𝑓 of π‘₯ on the closed interval must be equal to 12.

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