Absolute maxima and minima Determine the location and value of...

Question

Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.

ƒ(x) = x+ cos⁻¹x on [-1,1]

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to find the absolute maximum and minimum values of the function G of X, which is equal to 2 X plus the inverse sign of X on the closed interval from -1 to 1. We give 4 possible choices as our answers. For choice A, we have the absolute maximum value is the quantity of pi + 4 in quantity divided by 2, and the absolute minimum value is 2. For choice B, the absolute maximum value is the quantity of pi + 4 in quantity divided by 2, and the absolute minimum value is minus the quantity of pi + 4 in quantity divided by 2. For choice C, the absolute maximum value is 3, and the absolute minimum value is -2. And for choice D, the absolute maximum value is 2, and the absolute minimum value is minus the quantity of pi + 4 in quantity divided by 2. Now we're asked to find the absolute maximum and the absolute minimum values of our function G of X on the closed interval from -1 to 1. And so first we need to determine our critical points for our function G of X. To do that, I'll need to take a derivative of our function G with respect to X. So we need to calculate G of X. Which is going to be equal to the derivative with respect to X of the quantity of 2 X plus the inverse sine of X in quantity. So taking the derivative, this is going to be equal to. When I take the derivative of 2 X with respect to X, I have 2. And then we'll have plus the derivative with respect to X of the inverse sign of X. When I take that derivative, I get 1 divided by the square root of the quantity of 1 minus X squared in quantity. So, now we need to use our first derivative here to find the critical points. Recall that the critical points are located whenever our first derivative is equal to 0 or whenever the first derivative is undefined. So we'll look at the points in which our first derivative is undefined. And our first derivative is undefined when the denominator in our second term is equal to 0. So that means that the square root of the quantity of 1 minus X2d is going to be equal to 0 in order to find the location of our critical points. So we need to solve this equation for X. So first I'll square both sides of the equation, and I will get 1 minus X squared is equal to 0. I'll then add X2 to both sides, and so I'll get 1 is equal to X2. Then finally taking the square root of both sides, we will get X is equal to plus or minus 1. Now the other condition that we need to check is when our first derivative is equal to 0. So we will have 2 plus 1 divided by the square root of the quantity of 1 minus x 2 in quantity is equal to 0. So I'll subtract 2 from both sides of the equation. So I'll have 1 divided by square root of the quantity of 1 minus x 2 in quantity is equal to minus 2. Now, this equation has no solutions on our interval going from -1 to 1. So we have no solutions on the close interval from -1 to 1. So the only critical points that we need to evaluate happen to also be our end points. So we need to determine the value of G of X at X equal to -1 and X equals to + 1. So we'll start off with X equals to -1. So we'll have G of -1, and this is going to be equal to 2 multiplied by -1 plus the inverse sign of -1. This is going to be equal to -2, plus Minus pi divided by 2. So, we can simplify this expression by factoring out a minus sign, and then combining the remaining terms under a common denominator. So this is going to be equal to minus the quantity of pi + 4 in quantity divided by 2. So this is one of the values that we will use to determine whether we have an absolute maximum or absolute minimum. And then the other quantity that we need to calculate is G of 1. And so G of 1 is going to be equal to 2 multiplied by 1, plus the inverse sign of 1. And this is going to be equal to 2 plus pi divided by 2. And so combining those two terms with the common denominator, this is going to be equal to the quantity of pi plus 4 in quantity divided by 2. So that is the second value that we needed to evaluate. And so now I just need to determine which one is larger and which one is smaller, to determine whether we uh which one is the absolute maximum and which one is the absolute minimum. So, at X equal to -1, we see that we have G-1 is equal to minus the quantity of pi + 4 in quantity divided by 2, which is actually less than our value at G at 1. So this is going to be the absolute minimum. And that means that our other value. At G of 1, when X is equal to 1, which is the quantity of pi + 4 in quantity divided by 2, is going to be our absolute maximum. So the final answer to this problem is that we have The an absolute maximum value of the quantity of pi + 4 in quantity divided by 2, and we have an absolute minimum value of minus the quantity of pi plus 4 in quantity divided by 2. So that is our final answer for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.

ƒ(x) = x+ cos⁻¹x on [-1,1]

Key Concepts

Absolute Extrema

Critical Points

Endpoints of the Interval

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