Functions from derivatives Use the derivative f' to deter...

Question

Functions from derivatives Use the derivative f' to determine the x-coordinates of the local maxima and minima of f, and the intervals on which f is increasing or decreasing. Sketch a possible graph of f (f is not unique).

f'(x) = 10 sin 2x on [-2π, 2π]

f'(x) = 10 sin 2x on [-2π, 2π]

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says, given the derivative F of X is equal to 8 multiplied by cosine of X, minus 4 multiplied by the sine of 2 X on the close interval from 0 to 2 pi, identify the X coordinates of local maxima and minima of F, as well as the intervals where F is increasing or decreasing. Now, we want to identify the X coordinates where we have local maxima and minima, as well as determine the intervals over which our function is increasing or decreasing. To do this, we need to first find our critical points. Recall that our critical points are located whenever our first derivative is undefined, and here our first derivative is a combination of cosine and sine functions, both of which are going to be continuous and defined on the interval of 0 to 2 pi. But the other condition where we can have critical points is when the first derivative is equal to 0. So what we're going to do is set. 8, multiplied by cosine of X. -4, multiplied by sin of 2 X. Equal to 0. And we're gonna solve for X to find our critical points. So the first step to solve this for X is to rewrite our sine of 2 X in terms of sine and cosine using our double ang formulas. So we'll have 8 multiplied by cosine of X. -4, multiplied by the quantity of 2 multiplied by sin of X, multiplied by cosine of X in quantity, and that's going to be equal to 0. So since 4 multiplied by 2 is 8, I can factor out an 8 cosine of X out of each of these terms. We'll have 8 multiplied by cosine of X. Multiplied by the quantity of 1 minus sine of X. In quantity, and that's gonna be equal to 0. So to solve for X, I will set each one of these factors equal to 0. So, for the first factor we'll have 8 multiplied by cosine of X equal to 0. That means that cosine of X. Has to be equal to 0. And if I look at the values of X on the close interval from 0 to 2 pi, where this is true, we say that X is equal to 0, sorry, cosine of X is equal to 0, when X is equal to pi divided by 2, and 3 pi divided by 2. Now, if we look at the other factor, we have 1 minus sign of X, and I'll set that equal to 0 and solve for X. So that means that sign of X has to be equal to 1. And sine of X is equal to 1 when X is equal to pi divided by 2, which is the value that we already had calculated from the other factor. So our critical points are actually occurring when X is equal to pi divided by 2 and 3 pi divided by 2. We have two critical points that we need to check. So the next step is to divide our interval, going from 0 to 2 pi up into 3 smaller intervals that are broken up by our critical points here, and then we're going to use the first derivative test to determine whether our function F is increasing or decreasing on those intervals. So the first interval that we need to look at is going to be the open interval, going from 0 to pi divided by 2. And so, for the first derivative test, we're going to choose a point in this interval and calculate our first derivative and see whether it is positive or negative. So we're gonna choose X equal to pi divided by 4. So we'll look at F of pi divided by 4, which is going to be equal to 8, multiplied by the cosine of pi divided by 4. Minus Or multiplied by the sign. Of 2 multiplied by pi divided by 4. So, when we evaluate this expression, this is going to be equal to 4 multiplied by the square root of 2 minus 4. Which is greater than 0. So our first derivative here is positive, and we call that for the first derivative test that if our first derivative is positive, that means that our function F of X is going to be increasing. So our function is increasing on this interval. Now we'll look at the next interval. Which is going to be The interval going from. Pi divided by 223 pi divided by 2. So again, we'll choose a point in this interval to determine whether our function F of X is increasing or decreasing. So, we'll choose a point with X equal to pi, so we'll calculate F of pi. And that's gonna be equal to 8, multiplied by the cosine of pi. Minus 4 multiplied by the sign. Of Tupai. So when we evaluate this expression, this is going to be equal to -8. Which is less than 0. That means that our function is decreasing on this interval, since our first derivative here is negative. We call that for the first row test, that means that our function F of X is decreasing. And finally, for the last interval that we need to look at, it's going to be the open interval, going from 3 pi divided by 2, 22 pi. And here again, we'll choose a point in this interval, and we'll choose. X equal to 7 pi divided by 4, so we'll calculate F of 7 divided by 4. Which is going to be equal to 8 multiplied by the cosine. Of 7 pi divided by 4. -4, multiplied by the sign. Of 2 multiplied by 7 pi. Divided by 4. And when we evaluate this expression, this is going to be equal to 4 multiplied by the square root of 2, plus 4, which is greater than 0. That means that our function is increasing on this interval. So here we have identified the intervals over which our function F of X are increasing and decreasing. So now we can use this information to determine whether our critical points are local minima or maxima. So if we look at X equal to divided by 2, to the left of X equal to pi divided by 2, that interval is actually increasing, and just to the right of that point, that interval is actually decreasing. So if our function is going from increasing to decreasing, that means that we have a local maximum. So so we have a local. Maximum At X equal to pi divided by 2. If you look at X equal to 3 pi divided by 2. We see just to the interval just to the left of it, that that interval has our function F of X as decreasing, and to the interval just to the right of that, our function F of X is increasing. So if we have a function that is decreasing and then increasing, that gives us a local minimum. So we have a local. Minimum At X equal to 3 pi. Divided by 2. So those are the locations for our local Maxima and Mina. So just to recap of what we have here, we found that our critical points were located at X equal to pi divided by 2 and 3 pi divided by 2. We have a local maximum at X equal to pi divided by 2, a local minimum at X equals to 3 divided by 2, and we see that our function F of X. is increasing on the open interval from 0 to divided by 2, and the open interval from 3 divided by 2 to 2 pi. And our function F of X is decreasing on the open interval from pi divided by 2 to 3 divided by 2. So that here is the answer to this question. So I hope that this explanation has been useful, and I'll see you in the next video.

Functions from derivatives Use the derivative f' to determine the x-coordinates of the local maxima and minima of f, and the intervals on which f is increasing or decreasing. Sketch a possible graph of f (f is not unique).

f'(x) = 10 sin 2x on [-2π, 2π]

Key Concepts

Derivatives and Critical Points

Increasing and Decreasing Intervals

Graph Sketching

Watch next

Functions from derivatives Use the derivative f' to determine the x-coordinates of the local maxima and minima of f, and the intervals on which f is increasing or decreasing. Sketch a possible graph of f (f is not unique).f'(x) = 10 sin 2x on [-2π, 2π]

Key Concepts

Derivatives and Critical Points

Increasing and Decreasing Intervals

Graph Sketching

Watch next

Functions from derivatives Use the derivative f' to determine the x-coordinates of the local maxima and minima of f, and the intervals on which f is increasing or decreasing. Sketch a possible graph of f (f is not unique).

f'(x) = 10 sin 2x on [-2π, 2π]