Identifying Extrema

In Exerc...

Question

Identifying Extrema

In Exercises 41–52:

a. Identify the function’s local extreme values in the given domain, and say where they occur.

g(x) = x² − 4x + 4, 1 ≤ x < ∞

Accepted Answer

Welcome back, everyone. In this problem, we want to identify the local extreme values of the function H of Y equals Y2 minus 4Y + 4 in the domain where Y is greater than or equal to 1 but less than infinity. Now, if we're going to find the local extreme values, essentially we need to find the first derivative, find the critical points, that is where the first derivative is equal to zero, and then determine if either the critical points or end points are the or represent the local minimum value. So first, let's find the first derivative of H of Y. Now given each of Y. Equal to Y2 minus 4 Y + 4, OK. Then we know that the derivative of H of Y is going to be equal to Y minus 4. Now at the critical point. OK, at the critical point, then that means the first derivative is going to be equal to 0. So this implies that 2y minus 4 equals 0, which means that 2y will be equal to 4 and thus y will be equal to 2. So Y equals 2 is a critical point. And since it's within the given domain, then that means it's a candidate for a local extremal. Now we need to figure out if this is a local minimum or maximum. To do that, we can compute the second derivative, OK? And to find we can differentiate our first derivative again, OK? Now when we differentiate our first derivative again, then the second derivative of H of Y is going to be equal to 2. Since 2 is greater than 0 for all Y, then the function is, let's write that out, the function is concave up. Add Y equals to. And that means, therefore, Y equals 2 is a local minimum. Now that we know it's a low minimum, all we need to do is to evaluate the function to see what value of H it gives us, OK? Now, At Y equals 2, then H of 2 is going to be equal to 22 minus 4 multiplied by 2 + 4, which equals 4 minus 8 + 4, which equals 0. OK, so this is the value of H when Y equals 2. If we think about our end point here, if Y equals 1, then we know our function is going to be less or value is going to be less. So in that case then, OK, then that tells us. That the local minimum. OK, therefore, the local minimum. OK. Or rather Not that the value will be less, but the value will be more if Y equals 1. So we are sure that Y equals 2 is our local minimum. Therefore, our local minimum at Y equals 2. I 8 of 2, which equals 0. This is a local minimum for our function, and there is no local maximum. Thanks a lot for watching everyone. I hope this video helped.

Identifying Extrema

In Exercises 41–52:

a. Identify the function’s local extreme values in the given domain, and say where they occur.

g(x) = x² − 4x + 4, 1 ≤ x < ∞

Key Concepts

Local Extrema

Derivative and Critical Points

Quadratic Functions

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