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) = x² - 10 on [-2, 3]

Accepted Answer

Welcome back, everyone. On the interval open bracket -31 closed bracket, determine the absolute maximum and minimum values of the function F of X equals 2 X2 + 12 X + 20. A says the absolute maximum value is 34 at X equals 1, while the absolute minimum value is 2 at X E equals -3. B says it's the maximum is 34 at X equals 1 and the minimum is 20 at XE equals -3. C says the maximum is 20 at X equals 2 and the minimum is -2 at X equals -3, while B says the maximum is 14 at X equals 1, while the minimum is 2 at X equals -1. Now, to find the absolute maximum and minimum values, we can first find the critical points of our function by setting the derivative of FF X to 0. But if we're going to set it to zero, we need to know what its derivative is. So let's differentiate F of X. Now differentiating F of X. Then it's derivative. Is going to be equal to. 4 X + 12, OK. And like we said at the critical points. At the critical points. The derivative is equal to 0. So, if we set our derivative to 0, we can solve for the X value at our our values at our critical points. So in this case, This would mean that 4 X + 12 equals 0. So 4 X equals -12, which means X equals 12 divided by 4, which is -3. So this is the X value of our critical point for our graph. Or for our function. So in this case, Now that we have the critical point to find the absolute maximum and minimum values, we can evaluate. OK, evaluate FF X. At the critical points and the end points of our interval, OK? In other words, At the points where X equals, well, remember that our interval was -3 to 1 inclusive, so where X is -3 and X is 1. Well, one of the endpoints is already a critical point. So that means all we'd really be finding then is F of -3, the critical point and endpoint, and F of 1. Now F of -3 means we're gonna substitute X equals -3 into F of X, and that is gonna become 2 multiplied by -3 squad, plus 12 multiplied by -3 + 20, which is going to be equal to 2. On the other hand, when X E equals 1, F E equals 2 multiplied by 1 squared, plus 12 multiplied by 1 + 20. Which is equal to 34. Therefore, the absolute maximum value is 34 at X equals 1. OK. Let's, let's write that out. Because we notice here that's the highest value of F, OK. So the absolute maximum value. I 34. A X equals 1. And the absolute minimum value. The minimum value. Is 2 at X equals -3. Now if we compare that to our answer choices, then we can tell that option A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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) = x² - 10 on [-2, 3]

Key Concepts

Absolute Extrema

Critical Points

Closed Interval

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