Absolute Extrema on Finite Closed Intervals

Question

In Exercises 21–36, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.

g(x) = √(4 − x²), −2 ≤ x ≤ 1

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says determine the absolute minimum and maximum values of the function F of X, which is equal to the square root with quantity of 16 minus X squared in quantity on the closed interval from -1 to 4, and identify where they occur. So we need to determine the absolute minimum and maximum values of our function F of X on our given interval. So the first step we need to do is identify our critical points that will tell us the locations of possible local minima and maxima. So we need to take a derivative of function F of X, so we'll calculate F of X, and that's going to be equal to the derivative with respect to X. Of the square root of the quantity of 16 minus X squared in quantity. So taking this derivative, this is going to be equal to, and recall that when you take the derivative of the square root function, you're going to get 1 divided by the quantity of 2 multiplied by the square root of the quantity, which in this case is going to be 16 minus X squad in quantity. But we'll also need to apply our chain rules, so we need to multiply this fraction by the derivative with respect to X of 16 minus X2d. And applying our constant rule as well as our power rule for that, we'll get for that derivative minus 2 X. And so simplifying this expression, this is going to be equal to minus x divided by square root of the quantity of 16 minus X squad in quantity. So now recall to find critical points, those can be the locations at which our first derivative is either undefined or equal to zero. So let's look at the case when our first derivative is equal to 0, and that's going to occur when our numerator of our fraction is equal to 0. So in that case, we're going to have X equal to 0 as our first critical point. Now for the second critical point, we need to determine the X values at which our first derivative is undefined, and that's going to occur when the denominator of our fraction is equal to 0. So we're going to set the square root of the quantity of 16 minus X2 in quantity equal to 0. And so we need to solve this for X. So the first thing we'll do is square both sides of the equation. So we'll have 16 minus X2 is equal to 0. We'll then add X2 to both sides. We'll have 16 as equal to X2. Then taking the square root of both sides, we'll get X as equal to, and here we'd actually have plus or minus 4. However, -4 is outside the domain of our function, so we only need to include the X equal to 4 solution. So here we have two critical points. And what we need to do is determine whether we have a local minima or maxima at these critical points. Now, we don't need to do this analysis for X equal to 4, since that is also one of the endpoints of our interval. So we'll do a separate analysis on it. So what we're going to do is look to see if we have a local minimum or maximum at X equal to 0. And to do this, we need to determine whether we have a sign change of our first derivative at that point. And so, if you look at the case when X is less than 0, And if we look at our first root of F vex, the sign of F vex is determined entirely by the numerator, since the denominator being the square root of the quantity of 16 minus X squared in quantity is always going to be positive, since we're taking the positive square root. And so, if X is negative, we're gonna be multiplying that negative value by -1, and so they'll give us an overall positive quantity. So when X is less than 0, that means that a prime of X. It's actually going to be positive, so it's gonna be greater than 0. If you look at the case when X is greater than 0, Our X value is going to be positive, and when we multiply that by -1, that means that overall we're gonna have a negative value for our first derivative, so that means F of X. is going to be less than 0. So, we can see here that we do have a sign change of our first derivatives, since our first derivative is going from positive to negative at X equal to 0. And since we're going from positive to negative, that means that we have A local maximum at X equal to 0. So now what we need to do is calculate the value of F of X or X equal to 0, as well as the two endpoints of our interval. So we'll first calculate F of -1. Since that is our first endpoint, and so this is going to be equal to the square root. Of the quantity of 16 minus the quantity of minus 1 in quantity squared in quantity, and this is going to be equal to the square root of 15. Next, we'll look at F of 0, and that's going to be equal to the square root of the quantity of 16 minus 0 squared in quantity, and when we evaluate this expression, this is going to be equal to 4. Then finally, we need to calculate F of 4. And this is going to be equal to the square root of the quantity of 16 minus 4 squad in quantity, and evaluating this expression, this is going to be equal to 0. And so, for the absolute minimum, we're gonna need to choose the smallest value out of these 3 points. So, looking and comparing the values of our three points here, we see that we have an absolute minimum. That is equal to 0. At X equal to 4. And if we look for the absolute maximum, we'll do something similar, except we're looking for the largest value. Out of these 3 points, and 4 is slightly larger than the square root of 15, so that means the absolute maximum is equal to 4, and that occurs at X equal to 0. So, for our final answer, we'll have the absolute minimum is 0 at X equal to 4, and the absolute maximum is 4 at X equal to 0. So, I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Absolute Extrema

Critical Points

Graphing Functions

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