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

Determine the absolute maximum and minimum values of the function 𝑓(π‘₯) ={(6π‘₯ βˆ’ 3)Β², if π‘₯ ≀ 2 and 2 minus 9π‘₯, if π‘₯ > 2 in the interval [1, 6].

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Video Transcript

Determine the absolute maximum and minimum values of the function 𝑓 of π‘₯ is equal to six π‘₯ minus three squared, if π‘₯ is less than or equal to two and two minus nine π‘₯, if π‘₯ is greater than two in the closed interval one to six.

Remember, to find absolute extrema for continuous functions, we follow three steps. We begin by finding any critical points in the closed interval we’re looking at. We find the values of our function 𝑓 of π‘₯ at these critical points. And then, we check the end points for absolute extrema, values that are smaller than the relative minimum or larger than the relative maximum.

Now we have a slight problem here. This is a piecewise function. And we don’t yet know if this piecewise function is itself continuous. We do have a note that the function six π‘₯ minus three all squared is continuous, and the function two minus nine π‘₯ is continuous. So, what we’ll do is consider that π‘₯ equals two might be a critical point. And to test this, we’re going to evaluate the right-hand and left-hand derivative of our function at π‘₯ equals two.

We’ll begin by evaluating the right-hand derivative of 𝑓 at π‘₯ equals two. This is given by the limit as β„Ž approaches zero from the right of 𝑓 of two plus β„Ž minus 𝑓 of two all over β„Ž. We’re looking at the right-hand derivative, so we’re interested in the function that applies when π‘₯ is greater than two. That’s 𝑓 of π‘₯ equals two minus nine π‘₯.

So, we’re looking for the limit as β„Ž approaches zero from the right of two minus nine times two plus β„Ž minus two minus nine times two over β„Ž. That’s two minus 18 minus nine β„Ž minus two plus 18 all over β„Ž. This quite quickly simplifies to negative nine β„Ž over β„Ž. And then, we simplify further, and we see that we’re looking for the limit as β„Ž tends to zero from the right of negative nine. But this is independent of β„Ž, so we know that this is just going to be equal to negative nine. And so, our right-hand derivative is negative nine.

We’ll now repeat this process for the left-hand derivative. This time, we’re evaluating 𝑓 of two plus β„Ž minus 𝑓 of two over β„Ž as β„Ž approaches zero from the left. And so, we’re interested in the part of 𝑓 of π‘₯ where π‘₯ is less than or equal to two. So, we have the limit as β„Ž approaches zero from the left of six times two plus β„Ž minus three all squared minus six times two minus three all squared all over β„Ž. This simplifies to nine plus six β„Ž all squared minus nine squared over β„Ž.

And then, if we distribute the parentheses, we see that we’re left with the limit as β„Ž approaches zero from the left of 108β„Ž plus 36β„Ž squared. We can do a little bit of simplification, and this becomes 108 plus 36β„Ž. And then, we see that as β„Ž approaches zero from the left, we’re left with 108.

We can see that our left-hand and right-hand derivatives are not equal. And so, 𝑓 prime of π‘₯, our derivative, doesn’t actually exist at π‘₯ equals two. And so, we know that we have a critical point at π‘₯ equals two. And we know we’re going to need to find 𝑓 of π‘₯ at this point. We should also though check for any critical points on each part of our piecewise function. So, we’ll differentiate each part with respect to π‘₯ and set that equal to zero.

We can use the general power rule to differentiate six π‘₯ minus three all squared with respect to π‘₯. It’s two times six π‘₯ minus three. And then, we reduce the power by one. And then, we multiply that by the derivative of six π‘₯ minus three, which is just six. So, the derivative of this bit is 12 times six π‘₯ minus three. And the derivative of two minus nine π‘₯ is negative nine.

It should be quite clear that there’s no way for negative nine to be equal to zero. The derivative of this part of our function is always negative nine. But we can set 12 times six π‘₯ minus three equal to zero. And when we solve for π‘₯, we get π‘₯ equals 0.5. So, we have one more critical point at π‘₯ equals 0.5. We’re going to evaluate our function at the points π‘₯ equals two and π‘₯ equals 0.5 then. Let’s clear some space.

0.5 is less than two, so we evaluate the function at this point by using six π‘₯ minus three all squared. And when we substitute 0.5 in, we get zero. So, 𝑓 of 0.5 is zero. We use the same part of our piecewise function to evaluate 𝑓 of two. And when we do, we get 81. So, we’ve found 𝑓 of π‘₯ at the critical points on our function. Next, we need to check the end points.

These are 𝑓 of one and 𝑓 of six. We use six π‘₯ minus three all squared once again to evaluate 𝑓 of one. And that gives us nine. But six is greater than two. So, to evaluate 𝑓 of six, we use two minus nine π‘₯ and we get negative 52. We can see that the absolute maximum value of our function is 81 and the absolute minimum value is negative 52. The key point to remember here is that if we’re dealing with a piecewise function, we must check the behaviour at the end of each piece of our function. In our final example, we’ll consider how we can apply ideas about finding absolute extrema to exponential functions.

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