Values of related functions Suppose f is differentiable o...

Question

Values of related functions Suppose f is differentiable on (-∞,∞) and assume it has a local extreme value at the point x = 2, where f(2) = 0. Let g(x) = xf(x) + 1 and let h(x) = xf(x) + x +1, for all values of x.

a. Evaluate g(2), h(2), g'(2), and h'(2).

a. Evaluate g(2), h(2), g'(2), and h'(2).

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says let M be a differentiable function on the real numbers with a local maximum at X equal to 5, where M 5 is equal to -1. Define n of X is equal to 3x multiplied by m x + 5, and O of X is equal to 2 X multiplied by m of X + 3 X + 5. Calculate in of 50 of 5 in 5, and 05. Now, this problem wants us to calculate 4 different quantities, and the first quantity that we're going to calculate is in a 5. So we'll have N of 5 is equal to, and here we will use our expression for N of X. And so we'll have 3, multiplied by 5, since X is equal to 5, multiplied by M of 5. Plus 5. Now, we're told that M 5 is equal to -1. So this is going to be equal to 15, multiplied by -1. Plus 5. Which one we simplify this expression is going to be equal to -10. So, in the 5 is equal to -10, so that is our first quantity calculated. The next quantity that we will look at is 0 of 5. Which is equal to, and here we'll use our expression 40 X. And so that's going to be 2 multiplied by 5, since X is equal to 5, again, in this case, multiplied by M of 5. Plus 3 multiplied by 5. Plus 5. This is going to be equal to 10, multiplied by M of 5, which is -1. Plus 15 + 5. And when we supply this expression, this is going to be equal to 10. So, 05 is equal to 10. So that is the second quantity that we needed to calculate. Now, the third quantity that we need to calculate is in prime of 5. So, we need to first calculate in prime of X. So this is going to be equal to the derivative with respect to X. Of the quantity, and here we'll look at N of X, which is 3 X multiplied by m of X. Plus 5 So taking this derivative, this is going to be equal to. We take the derivative of the first termin or sum, we're going to need to use our product rule. So we call for the product rule that we take the derivative of the first termin of our product, so we'll take the derivative with respect to X of 3 X, which is 3, that gets multiplied by the second terminal product, which is M of X. Then we'll have plus the first term in a product, which is 3X, multiplied by the derivative of the second term in a product. So we'll have the derivative with respect to X of M of X, which is going to be M X. Then we'll have plus the derivative of 5 with respect to X, and here we're taking a derivative of a constant, so it's going to be equal to 0, so we'll have plus 0 as the last term. So now that we have an equation for in of X, we can now calculate in prime of 5. Which is going to be equal to 3, multiplied by M of 5. Plus 3 multiplied by 5, multiplied by M of 5. And here we need to determine the value for M 5, and we're told that we have a local maximum at X equal to 5. And so we have a local maximum, that means that our first derivative is 0. So that means that M 5 is equal to 0. So that means that N of 5 is going to be equal to 3 multiplied by M of 5, which is -1. Plus 15 multiplied by M of 5, which is 0. And when we evaluate this expression, this is going to be equal to -3. So, in 5 is equal to -3. That is the third quantity that we needed to calculate. Now for the 4th quantity, we need to calculate of 5. And here we need to calculate O of X first. So this is going to be equal to the derivative with respect to X of the quantity of 2 X multiplied by M of X. Plus 3 multiplied by X. Plus 5 in quantity. So taking the derivative, this is going to be equal to. For the first term in our sum, we're again going to have to use the product rule. So I'll take the derivative of the first term with respect to X, which is for the first term 2 X. So when I take that derivative, I will get 2, and that gets multiplied by the second term in our product, which is M of X. Then we'll have plus the first term in a product, which is 2 X, multiplied by the derivative with respect to X of the second term in our product, which is going to be M of X. Then we'll have less, the derivative with respect to X of 3 X, which is 3. Plus the derivative of 5 with respect to X, which is 0. So now that we have an equation for O of X, we can calculate of 5. So of 5. is equal to. 2 multiplied by M of 5. Plus 2 multiplied by 5, multiplied by M of 5. Plus 3. So this is going to be equal to 2, multiplied by M 5, which is minus 1. Plus 10 multiplied by M5, which is 0 + 3. So some find this expression, this is just going to be equal to 1. So 5 is equal to 1. And so that is the last quantity that we need to calculate for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Values of related functions Suppose f is differentiable on (-∞,∞) and assume it has a local extreme value at the point x = 2, where f(2) = 0. Let g(x) = xf(x) + 1 and let h(x) = xf(x) + x +1, for all values of x.

a. Evaluate g(2), h(2), g'(2), and h'(2).

Key Concepts

Differentiability and Local Extrema

Function Evaluation

Derivative of a Product

Watch next

Values of related functions Suppose f is differentiable on (-∞,∞) and assume it has a local extreme value at the point x = 2, where f(2) = 0. Let g(x) = xf(x) + 1 and let h(x) = xf(x) + x +1, for all values of x.a. Evaluate g(2), h(2), g'(2), and h'(2).

Key Concepts

Differentiability and Local Extrema

Function Evaluation

Derivative of a Product

Watch next

Values of related functions Suppose f is differentiable on (-∞,∞) and assume it has a local extreme value at the point x = 2, where f(2) = 0. Let g(x) = xf(x) + 1 and let h(x) = xf(x) + x +1, for all values of x.

a. Evaluate g(2), h(2), g'(2), and h'(2).