Each of Exercises 67–88 gives the first derivative of a contin...

Question

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.
80. y' = 1 - cot²θ, for 0 < θ < π

80. y' = 1 - cot²θ, for 0 < θ < π

Accepted Answer

Below there today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. The first derivative of a continuous function F of X is F of X is equal to 1 minus tangent square of X on divided by 23 divided by 2. Find the second derivative of F of X and sketch the general shape of its graph. Awesome. So it appears for this particular problem we're asked to solve for two separate answers. Our first answer is we're asked to determine or to find the second derivative of this function of F of X. And we're also asked to sketch the general shape of its graph of this particular graph of the second derivative. So that's our second and final answer we're ultimately trying to solve for. And we're also given a blank coordinate plane in order to draw our sketch. That's what this box represents. So what we need to do first off in order to solve this particular problem is we need to differentiate the first derivative with respect to X in order to determine the second derivative. So, with that said, we can say that F of X, which that means the second derivative of F of X. is equal to D by DX 1 minus tangent square of X. Awesome. So when we begin to perform our derivation with respect to X, we will find that it's equal to -2, multiplied by tangent of X, multiplied by D by DX of tangent of X, which is going to be equal to -2 multiplied by tangent. Of X multiplied by C can squad of X. Fantastic. So, let us also quickly recall, as we should recall, let us note and recall that tangent of X is equal to sin of x divided by cosine of X. Awesome. So with that said, so we can also write, or we need to note that we could also write our second derivative as F of X, which is the second derivative of F of X is equal to negative2 multiplied by sin of x divided by cosine of X. Multiplied by one divided by Cosine squared of X, which will be all equal to negative. 2, so 2 multiplied by sign of X. Divided by cosine. To the power of 3 of X. Awesome. So we can write that F of X is equal to -2 multiplied by sin of x divided by cosine of X, multiplied by 1 divided by cosine square of X, which is also equal to -2 multiplied by sin of x, all divided by cosine to the power of 3 of X. Fantastic. So with that said, in order to sketch the graph of F of X, we need to determine the critical and inflection points. So as we should recall, a critical point occurs where the first derivative is zero or undefined. So with that in mind, we need to find where the first derivative is undefined. So let us note that since. Tangent squared of X is equal to squad of x divided by cosine squared of X, we can go ahead. And safely write that F of X is equal to 1 minus 2 X. So sin squared of X. And then we're gonna take sin squared of X divided by cosine squared of X, which is going to be equal to cosine quad x minus squad of X, all divided by cosine squared of X. Fantastic. So, moving right along here. So the derivative will be undefined when, let's make a note of this, it will be undefined when. Cosine squared of X is equal to 0, so that means that cosine squared of X is equal to 0. Which will mean that cosine of X is equal to 0. So that means that the general solution is that X is equal to pi divided by 2 plus N multiplied by pi. Awesome. So let us also note that there is no value of X on parentheses divided by 2.3 divided by 2, where the first derivative is undefined. So with that in mind, we need to set the first derivative equal to 0, so we need to take 1 minus. Tangent squared of X is equal to 0, so then we can say that tangent squared of X is equal to 1. So that will mean that tangent of X is simply equal to plus or minus 1. So thus, therefore, the general solution for tangent of X is equal to 1. So let us make a note. So. Tangent of X is equal to 1. For this condition, it will state that X is equal to pi divided by 4 plus N multiplied by pi. Um and this is for where N is an element of Z. Awesome. So or Z, so it's like an element. Of like all real numbers, it's a fancy way of writing it. So, moving right along here, so the valid solution for the given interval as a result will be that X is equal to 5 pi divided by 4. So now, moving our attention to solve for the condition where tangent of X is equal to -1 now, that will mean that X is equal to 3 divided by 4 plus N multiplied by pi, where N is an element of Z once again. So that will mean that the valid solution for this particular interval will be that X is equal to 3 pi divided by 4. So that will mean as a result, that our critical points will be, and I'll write them in blue, so our critical points, I'll call them CP for short. So our critical points will be X is equal to. 3 pi divided by 4 and X is equal to 5 pi divided by 4. Fantastic. So, moving right along here. So now, our next step is we need to note that these points will split the given interval into three separate intervals. So that will mean that we have parentheses divided by 23 divided by 4. Once again in parentheses, parentheses. 3 divided by 45 divided by 4, and then finally, parentheses, 5 pi divided by 43 pi divided by 2. Fantastic. So, moving right along here. So now we need to recall and use the first derivative test to determine whether X equals 3 pi divided by 4 and X equals 5 divided by 4 represent a local maximum, minimum, or neither. So the first derivative test, as we should recall, involves analyzing the sign of the derivative on either side of the critical point. So, as we should recall, if the derivative changes from positive to negative, it is a local maximum. If it changes from negative to positive, it is a local minimum, and if there is no change, it is neither a maximum nor minimum. This is also called a saddle point, if that is the case. So, with that said, for the condition parentheses pi divided by 2, 3 pi divided by 4. Let us say that X is equal to 2 divided by 3. So 2 pi divided by 3. So focusing on this condition now, you can go ahead and write that 1 minus squared of 2 pi divided by 3. Divided by cosine squared of 2 pi divided by 3. Is going to be equal to so when we plug this big old long expression into our calculator, we will find that it's simply equal to -2. And because it's negative, that means that F of X is decreasing, so I have an arrow pointing down. So now focusing on our other condition, which is parentheses 3 pi divided by 4. 5 pi divided by 4 in parentheses, and let us say that X is equal to pi in this case. And once again, when I say condition, this is the interval we're paying attention to. So when we plug in a value of pi in for X, so instead of rewriting out the entire formula again, so we're using our 1 minus squad of X divided by cosine squared of X, we're plugging in a value of pi in for x. So that will mean for this particular condition, when we plug in a value of pi in for X, we will find that it's going to be simply equal to 1. Which will mean that it is increasing as it's a positive value. So that means that F of X is increasing. So, finally, for our last interval, which is parentheses 5 pi divided by 4.3 pi divided by 2. We're going to let X equal 4 divided by 3 in this particular case. So when we plug in a value of X being 4 pi divided by 3, we will find that it's going to be simply equal to -2, and because it's a negative value, that means that F of X is decreasing. So that means, without a doubt that at X equals 3 pi divided by 4, the sign of the derivative changes from negative to positive. Therefore, it is a local minimum. So once again, so when X is equal to. 3 pi divided by 4. So let's make a note of this. This isn't important. So we have a local min, which I'll call it L min, is equal to. So we're gonna have a local min rather than saying it's equal to. So we have our local minimum at X equals 3 pi divided by 4. And we have a local maximum, which I'll call L max. We have a local maximum at X equals 5 pi divided by 4, and this is because the sin of the derivative changes from positive to negative, which makes it a local maximum. And once again, our local minimum, which is 3 pi divided by 4, because there's a sign of the derivative changes from negative to positive, it's a local minimum. Now our next step at this point is we need to determine the inflection points. So as you should recall, an inflection point occurs where the second derivative is 0 or undefined, so we need to set the second derivative equal to 0, so we need to take -2 multiplied by sin of x divided by cosine to the power of 3 of X is equal to 0. So that will mean that the fraction will be zero when, let's make a note of this, when. Sign of X is equal to 0, so that means that the general solution for sin of X is equal to 0 will be that X is equal to N multiplied by pi for when N is an element of Z. Fantastic. So, that means that the valid solution for the given interval will be simply X is equal to pi. So now at this point, we need to find where F of X is undefined, so the fraction will become undefined when the denominator is 0. So that means we need to take cosine to the power of 3 of X is equal to 0. So that will mean that cosine of X is equal to 0. And because cosine of X is equal to 0, we can simply state that the general solution is X is equal to pi divided by 2. Plusn multiplied by pi. So that means as a result, there is no value of X on parentheses pi divided by 2.3 pi divided by 2, where the second derivative is undefined. We also need to recall a note that the inflection point in this particular case, which is the given interval, into the two intervals for the test for concavity, so we will have Once again, the inflection point will split the given interval into two intervals to help us test for concavity, so we need to take parentheses. Pi divided by 2, and parentheses 3 pi divided by 2, once again in parentheses. So, with that said, we need to determine the concavity of F of X at. So once again, we're trying to find, I'll write this in blue. So F of X. At X equals pi. So we need to note that F of X, so let's make a note of this. So F of X, so F of X is gonna be concave up, which I'll call Cu, so it's gonna be concave up when F of X is greater than 0, and it's going to be concave down, which I'll call CD for concave down. So it's going to be concave down win, drum roll. F of X is less than 0. Fantastic. So, with that said, Our next step now is we need to focus our attention on The condition for pi divided by 2, in parentheses. Let us state that X is going to be equal to 5 pi divided by 6. So that will mean we will plug in a value of X equals 5 divided by 6 into our function. So we will take -2 multiplied by sin of 5 pi divided by 6 divided by cosine to the power of 3 of. 5 pi divided by 6. So when we plug in this big old long expression into our calculator, we will find that it's simply equal to 8 multiplied by the square root of 3 divided by 9, which is a positive value, so that will mean that it's going to be concave up as a result. Awesome. So now, our next step is we need to consider our other condition for the interval, so we need to take parentheses. Pi 3 pi divided by 2, and let us say that X is equal to 7 pi divided by 6 in this case. And instead of rewriting our expression out again, I'll just tell you what it's simply equal to. So when you plug in a Value of X equals 7 pi divided by 6 into our function, we will find that it's equal to -8 multiplied by the square root of 3 divided by 9, and because it's a negative value, it's going to be concave down as a result. So that means we can now sketch the general shape of our graph for F of X. So we need to know once again that 4 parentheses pi divided by 23 divided by 4. It's going to be decreasing concave up. We have a local minimum at X equals 3 pi divided by 4, and for parentheses 3 divided by 4 pi. It's increasing concave up. We have an inflection point at X equals pi and 4 parentheses 5 divided by 4. It's increasing concave down. We also have a local maximum at X equals 5 divided by 4. And finally, for 5 pi divided by 43 pi divided by 2, it will be decreasing concave down. So with that said, Let us go and sketch a graph for our function based on all the information that we solve for together. Since now we have a generalization of the shape of our sketch should be. So we now know what our sketch should be after we performed all that math together. So with that said, We should produce the following graph. Or rather a graph of our sketch. So our final answers without a doubt have to be for first off, our second derivative is F of X is equal to -2 multiplied by tangent of X multiplied by C2 X. And for our sketch of the curve, it looks to appears to be cubic-like in nature. It starts out pointing upwards and then it curves on the shape of a parabo, and then it starts to curve up, then it dips down to like an inverted parabola shape, and then stretches downwards. So our curve starts pointing upwards and then it curves down, curves up a little bit, and then curves down. And we have it at X equals 3 pi divided by 4, X equals pi, and then X equals 5 pi divided by 4. And that's it. We've officially solved for this problem. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.
80. y' = 1 - cot²θ, for 0 < θ < π

Key Concepts

First Derivative

Second Derivative

Graphing Procedure

Watch next

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.80. y' = 1 - cot²θ, for 0 < θ < π

Key Concepts

First Derivative

Second Derivative

Graphing Procedure

Watch next

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.
80. y' = 1 - cot²θ, for 0 < θ < π