Identifying Extrema

In Exerc...

Question

Identifying Extrema

In Exercises 53–60:

a. Find the local extrema of each function on the given interval, and say where they occur.

f(x) = sec²x − 2tan x, −π/2 < x < π/2

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says determine the local extrema of GFT, which is equal to 22 of T, plus 2 multiplied by tangent of T. 4 minus pi divided by 2 is less than T, which is less than pi divided by 2. So we need to determine the local extrema of GFT on the given interval and the problem. So, in order to find the local extrema, we first need to find the critical points, and to find the critical points, we need the derivative of G with respect to T. So we need to calculate G of T. And that's going to be equal to the derivative with respect to T. Of the quantity of 2 squared of T. Plus 2 multiplied by tangent of T. In quantity So this, taking this derivative, this is going to be equal to, when we take the derivative of the first term, we're gonna have to apply our power rule as well as our chain rule. So applying the power rule, we'll get 2 multiplied by sequin of T. They will need to apply a chain rule and multiply this by the derivative with respect tot of sequin of T, and recall that that derivative is equal to sequent of T multiplied by tangent of T. Then we'll have plus 2 multiplied by the derivative with respect tot of tangent to T for the second term, and when we take that derivative, we get the squared of T. So, we can simplify this expression by factoring out a 2 multiplied by squared of T out of both terms. In doing so, this is going to be equal to 2 multiplied by 2 squared of T. Multiplied by the quantity of tangent of tea. Plus one in quantity. So now we need to find our critical points and recall that critical points occur whenever our first derivative is either undefined or equal to zero. Now if we look at our function here, we have both a deacon function and a tangent function, and both of those functions are undefined at plus or minus pi divided by 2. However, those two points fall outside of the domain of a function of GFT. So we don't need to worry about them, since they can't be critical points since they're not included in the domain. So, the only thing we need to do is set our first derivative equal to 0 and solve for T. So we're going to set 2 multiplied by squared of T, multiplied by the quantity of tangent of T. Plus 1 in quantity equal to 0. Now, see of T can never be equal to 0 on the interval that we have given in the problem, so that means that the second term in that product has to be equal to 0. So we're going to set tangent of T. Plus one Equal to 0 and solve for T. So we'll start off by subtracting one from both sides. So we'll have the tangent of T equal to -1, and taking the inverse tangent of both sides, we'll get T is equal to minus pi divided by 4. Now tangent of T can take on, uh, can take the value of -1 at different values oft other than minus pi divided by 4. However, that is the only value of T that falls within our domain. So we need to check to see if this critical point is a local minimum or maximum. So for that, we'll use the second derivative test. So we need to calculate G of T, and that's going to be equal to the derivative with respect to T of the quantity of 2 multiplied by squared of T, multiplied by the quantity of tangent of T. Plus one in quantity. And so here we need to take this derivative, and we're taking a derivative of a product. So we're going to need to apply our product rule. So this is going to be equal to, and here we call your product rule. That is that we're going to have the derivative of the first term of products. So we'll have the derivative with respect tot of 2 multiplied by sequence square of T. So we'll have 2 multiplied by the derivative with spect of squared T. We found that earlier, and that's going to be the quantity of 2 multiplied by sequent squared of T, multiplied by tangent of T. In quantity and that gets multiplied by the second term in a product which is the quantity of tangent of tea. Plus 1 in quantity. Then we'll have plus the first term of a product which is 2 multiplied by sin squared of T. And that gets multiplied by the derivative with spect of the second term of a product. So that's gonna be the derivative with respect to T of quantity of tangent of T plus 1 in quantity. When we take that derivative, that's going to be equal to sequent squared of T. So, simplifying this expression, this is going to be equal to. 4 multiplied by 2 squared of T, multiplied by tangent of T. Multiplied by the quantity of tangent of T plus 1 in quantity, plus 2 multiplied by 2 to the 4th of T. And so now we just need to evaluate this expression when T is equal to minus pi divided by 4. So calculating G of minus pi divided by 4. This is going to be equal to 4 multiplied by squared of minus pi divided by 4. Multiplied by tangent of minus pi divided by 4. Multiplied by the quantity of tangent of minus pi divided by 4. Plus 1 in quantity, plus 2 multiplied by sequent to the 4th of minus pi divided by 4. So recall that the sequent of minus pi divided by 4 is equal to square root 2, and the tangent of minus pi divided by 4 is equal to -1. And so evaluating this expression, this is going to be equal to 16. So here we see that our second derivative is positive at our critical point. So that means that we have a local. Minimum at this point. So we have a local minimum at T equal to minus pi divided by 4. So now all we need to do is calculate the value of our function G at this point. We'll calculate G of minus pi divided by 4, and that's going to be equal to sequin squad. Of minus pi divided by 4, plus 2 multiplied by tangent of minus pi divided by 4. And if I in this expression, this is going to be equal to 0. So now we have all the information that we need to write our final answer. So, for a final answer, we'll have a local minimum. At T equal to minus pi divided by 4, with G of minus pi divided by 4 equal to 0. And if we were to look at local maxima, we would see that we have none, since we only found a local minimum. So these are the answers that the problem was actually looking for. So I hope that this explanation has been useful, and I'll see you in the next video.

Identifying Extrema

In Exercises 53–60:

a. Find the local extrema of each function on the given interval, and say where they occur.

f(x) = sec²x − 2tan x, −π/2 < x < π/2

Key Concepts

Local Extrema

First Derivative Test

Trigonometric Derivatives

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