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 − 6x − 9,−4 ≤ x < ∞

Accepted Answer

Welcome back, everyone. In this problem, we want to identify the local extreme values of the function P X equals -3 X2 + 12 X + 11 in the domain where X is greater than or equal to 3 but less than infinity. Now to find the local extreme values we'll need to differentiate PF X, set it to 0 to get the critical points, and then test the nature of those critical points whether they are maximum or minimum. So first, let's start by differentiating P X. Now we know that PFX is equal to -3 X2 + 12 X + 11. So that means that the derivative of PFX is going to be equal to -6 X + 12. Now at the critical point. We know that the derivative of PF X is going to be equal to 0. So this implies then that -6 X + 12 will be equal to 0, OK. Which means then that 6 X will be equal to 12 and thus X will be equal to 2. Therefore, X equals 2 is a critical point of a function p of X. And since it's within the given domain between uh that is open bracket -3 to infinity in parentheses, that means it is a candidate for a local extremal. So now we need to test if it is a local minimum or maximum. And we can use that by the 2nd derivative test. No, by the second derivative test, what we mean is that we're going to differentiate our function twice, OK? Or in other words, we'll differentiate the derivative of PFX, OK? And once we have the second derivative, if it is negative, that means the function is concave done and it's a local maximum. But if it is positive, that means the function is concave up and it's a local minimum. Now we know the first derivative of -6 X + 12, so that means the second derivative. Is going to be equal to -6. 6 is less than 0, OK. So thus, that means the function is concave down. At X equals 2, which means that X equals 2 is a local maximum. Now that we know that's the X value for our maximum, then let's find the value of our function. Now, this means then that we'll need to find P2. Which is gonna be -3 multiplied by 22 + 12 multiplied by 2 + 11. 2 squared is 4, -3 multiplied by 4 is -12, 12 multiplied by 2 is 24, and -12 + 24 + 11 equals 23. Therefore, this tells us. OK, that the local maximum. Local maximum. Is at X equals 2 with a value of 2 being equal to 23 and there is no local minimum in the domain. Thanks 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 − 6x − 9,−4 ≤ x < ∞

Key Concepts

Local Extrema

Derivative

Quadratic Functions

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