# Video: Finding the Absolute Maximum and Minimum Values of a Piecewise Function in a Given Interval

In the interval [−1, 2], determine the absolute maximum and minimum values of the function 𝑓(𝑥) = 4𝑥² + 3𝑥 − 7 if 𝑥 ≤ 1 and 𝑓(𝑥) = 6𝑥 − 5 if 𝑥 > 1, and round them to the nearest hundredth.

03:15

### Video Transcript

In the closed interval negative one to two, determine the absolute maximum and minimum values of the function 𝑓 of 𝑥 equals four 𝑥 squared plus three 𝑥 minus seven if 𝑥 is less than or equal to one, six 𝑥 minus five if 𝑥 is greater than one. And round them to the nearest hundredth.

In order to find absolute maximum and minimum values of a function, we need to consider a couple of things. We need to look for any critical points of the function. Remember, these are points where the first derivative is equal to zero. A critical point can indicate a relative maximum and a relative minimum. And these of course could also be absolute extrema. We also need to consider the end points of the function.

So what we’re going to do is begin by finding the first derivative of our function. Now, our function is piecewise. So we’re going to differentiate each expression with respect to 𝑥. Let’s begin with the expression four 𝑥 squared plus three 𝑥 minus seven. The derivative of four 𝑥 squared with respect to 𝑥 is two times four 𝑥, which is eight 𝑥. The derivative of three 𝑥 is three. And the derivative of negative seven is zero. And that’s for values of 𝑥 less than or equal to one.

Next, we differentiate six 𝑥 minus five with respect to 𝑥. Well, the derivative of six 𝑥 minus five is simply six. So we see that the first derivative is either equal to 𝑥 plus three if 𝑥 is less than or equal to one or six if 𝑥 is greater than one. And actually, if we look carefully, we see that the expression six cannot be equal to zero for any values of 𝑥. This indicates to us there are no critical points in this part of the function. But we will set eight 𝑥 plus three equal to zero and solve for 𝑥. By subtracting three from both sides and then dividing through by eight, we find that 𝑥 equals negative three-eighths. So we have a critical point when 𝑥 is equal to negative three-eighths.

So we’re going to do a number of things. We’re going to evaluate the function at this critical point. That’s 𝑓 of negative three-eighths. We’ll evaluate it at the lower and upper end of our interval. That’s 𝑓 of negative one and 𝑓 of two. And we’ll evaluate it here when 𝑥 is equal to one. This is the end point of each bit of the function. Now, when 𝑥 is equal to negative three-eighths, it is indeed less than or equal to one. So we use the first bit of our function to evaluate 𝑓 of negative three-eighths. It’s four times negative three eighths squared plus three times negative three-eighths minus seven. That’s negative 7.5625.

When 𝑥 is equal to negative one, we use the same bit of the function. So we get four times negative one squared plus three times negative one minus seven, which is negative six. Now, negative six is greater than negative 7.5625. So when 𝑥 is equal to negative one, we absolutely can’t have an absolute minimum. But we could still have an absolute maximum. We’re now going to test 𝑓 of two. When 𝑥 is equal to two, this is obviously greater than one. So we’re going to use the second part of our function. That’s six times two minus five, which is equal to seven.

Finally, we just need to check when 𝑥 is equal to one. We go back to the first part of our function because this is defined for values of 𝑥 less than or equal to one. We get four times one squared plus three times one minus seven, which is equal to zero. Well, if we look carefully, we see that the smallest value we have over our interval is negative 7.5625. And the largest value we get is seven. We round negative 7.5625 correct to the nearest hundredth.

And we find that the absolute maximum value of our function is seven. And the absolute minimum is negative 7.56.