2011 Calculus AB free response #4b

Part b. "Determine the x-coordinate of the point at which g has an absolute maximum on the interval negative 4 is less than or equal to x, is less than or equal to 3. And justify your answer." So let's just think about it in general terms. If we just think about a general function over an interval, where it could have an absolute maximum. So let me draw some axes over here. And I'm speaking in general terms first, and then we can go back to our function g, which is derived from this function f right over here. So let's say that these are my coordinate axes, and let's say we care about some interval here. So let's say this is the interval that I care about. A function could look something like this. And in this case, its absolute maximum is going to occur at the beginning of the interval. Or a function could look something like this. And then the absolute maximum could occur at the end point of the interval. Or the other possibility is that the function looks something like this. At which point, the maximum would be at this critical point. And I say critical point, as opposed to just a point where the slope is zero, because it's possible to the functions not differentiable there. You could imagine a function that looks like this, and maybe wouldn't be differentiable there. But that point there still would be the absolute maximum. So what we really just have to do is evaluate g at the different endpoints of this interval, to see how high it gets, or how large of a value we get for the g at the end points. And then we have to see if g has any critical points in between. And then evaluate it there to see if that's a candidate for the global maximum. So let's just evaluate g of the different points. So let's start off, let's evaluate g at negative 4-- at kind of the lowest end, or the starting point of our interval. So g of negative 4 is equal to 2 times negative 4 plus the integral from 0 to negative 4 f of t dt. The first part is very easy, 2 times negative 4 is negative 8. Let me do it over here so I have some real estate. So this is equal to negative 8. And instead of leaving this as 0 to negative 4 f of t dt, let's change the bounds of integration here. Especially so that we can get the lower number as the lower bound. And that way, it becomes a little bit more natural to think of it in terms of areas. So this expression right here can be rewritten as the negative of the integral between negative 4 and 0 f of t dt. And now this expression right over here is the area under f of t, or in this case, f of x-- or the area under f between negative 4 and 0. So it's this area right over here. And we have to be careful, because this part over here is below the x-axis. So this we would consider negative area when we think of it in integration terms. And this would be positive area. So the total area here is going to be this positive area minus this area right over here. So let's think about what this is. So this area-- This section over here we did this in part a, actually, This section-- this is a quarter circle, so it's 1/4-- so these are both quarter circles. So we could multiply 1/4 times the area of this entire circle, if we were to draw the entire thing all the way around. It has a radius of 3. So the area of the entire circle would be pi times 3 squared or it would be 9 pi. And of course we're going to divide it by 4-- multiply it by 1/4 to just get this quarter circle right over there. And then this area right over here, the area of the entire circle, we have a radius of 1. So it's going to be pi times 1 squared. So it's going to be pi, and then we're going to divide it by four, because it's only one fourth of that circle. And we're going to subtract that. So we have negative pi and we were multiplying it times 1/4 out here, because it's just a quarter circle in either case. And we're subtracting it, because the area is below the x-axis. And so this simplifies to-- this is equal to 1/4 times 8 pi, which is the same thing as 2 pi. Did I do that right? 1/4 times 8 pi. So this all simplifies to 2 pi. So g of negative 4 is equal to negative 8 minus 2 pi. So clearly it is a negative number here. More negative than negative 8. So let's try the other bounds. So let's see what g of positive 3 is. I'll do it over here so I have some more space. So g of positive 3, when x is equal to 3, that-- we go back to our definition-- that is 2 times 3 plus the integral from 0 to 3 f of t dt. And this is going to be equal to 2 times 3 is 6. And the integral from 0 to 3 f of t dt, that's this entire area. So we have positive area over here. And then we have an equal negative area right over here because it's below the x-axis. So the integral from 0 to 3 is just going to be 0, you're going to have this positive area, and then this negative area right over here is going to completely cancel it out. Because it's symmetric right over here. So this thing is going to evaluate to 0. So g of 3 is 6. So we already know that our starting point, g of negative 4-- that when x is equal to negative 4-- that is not where g hits a global maximum. Because that's a negative number. And we already found the end point, where g hits a positive value. So negative 4 is definitely not a candidate. x is equal to 3 is still in the running for the x-coordinate where g has a global absolute maximum. Now what we have to do is figure out any critical points that g has in between. So points in there where it's either undifferentiable or its derivative is equal to 0. So let's look at this derivative. So g prime of x-- we just take the derivative of this business up here. Derivative of 2x is 2. Derivative of this definition going from 0 to x of f of t dt-- we did that in part a, this is just the fundamental theorem of calculus-- this is just going to be plus f of x right over there. So it actually turns out that g is differentiable over the entire interval. You give any x value over this interval, we have a value for f of x. f of x isn't differentiable everywhere, but definitely f of x is defined everywhere, over the interval. So you'll get a number here, and obviously two is just two, and you add two to it, and you get the derivative of g at that point in the interval. So g is actually differentiable throughout the interval. So the only critical points would be where this derivative is equal to 0. So let's set this thing equal to 0. So we want to solve the equation-- I'll just rewrite it actually-- So we want to solve the equation g prime of x is equal to 0, or 2 plus f of x is equal to 0. You can subtract 2 from both sides, and you get f of x is equal to negative 2. So any x that satisfies f of x is equal to negative 2 is a point where the derivative of g is equal to 0. And let's see if f of x is equal to negative 2 at any point. So let me draw a line over here at negative 2. We have to look at it visually, because there's only given us this visual definition of f of x. Doesn't equal negative 2, doesn't equal to negative 2, only equals negative 2 right over there. And it looks like we're at about 2 and 1/2, but let's get exact. Let's actually figure out the slope of the line, and figure out what x value actually gives f of x equal to negative 2. And we could figure out the slope of this line fairly visually-- or figure out the equation of this line fairly visually, we can figure out its slope. If we run 3-- if our change in x is 3, then our change in y, our rise, is negative 6. Change in y is equal to negative 6. Slope is rise over run, or change in y over change in x. So negative 6 divided by 3 is negative 2. It has a slope of negative 2. And actually, I could have done that easier. Where if we go forward one, we go down by 2. So we could have seen that the slope is negative 2. So this part of f of x, we have y is equal to negative 2x plus-- and then the y-intercept is pretty straightforward. This is at 3-- 1, 2, 3. Negative 2x plus 3. So part of f of x where clearly it equals negative 2 at some point of that-- this part of f of x is defined by this line. Obviously this part of f of x is not defined by that line. But to figure out the exact value, we just have to figure out when this line is equal to negative 2. So we have negative 2x plus 3 is equal to negative 2. And remember, this isn't-- this is what f of x is equal to, over the interval that we care about. If we were talking about f of x over there, we wouldn't be able to put a negative 2x plus 3, we would have to have some form of equation for these circles. But right over here, this is what f of x is, and now we can solve this pretty straightforwardly. So we can subtract 3 from both sides, and we get negative 2x is equal to negative 5. Divide both sides by negative 2, you get x is equal to negative 5 over negative 2, which is equal to 5/2. Which is exactly what we thought it was when we looked at it visually. It looked like we were at about 2 and 1/2, which is the same thing as 5/2. Now we don't know what this is. We don't know if this is an inflection point. Is this a maximum? Is this a minimum? So really we just want to evaluate g at this point to see if it gets higher than when we evaluate g at 3. So let's evaluate g at 5/2. So g at 5/2 is going to be equal to 2 times 15/2 plus the integral from 0 to 5/2 of f of t dt. So this first part right over here, the 2's cancel out. So this is going to be equal to 5. And then plus the integral from 0 to 5/2. Now, you might be able to do it visually, but we know what the value is of f of t over this interval, we already figured out the equation for it over this interval. So it's the integral of negative 2x plus 3 dt. And then let's just evaluate this. Let me get some real estate. So this is-- let me draw a line here., so we don't get confused. So this is going to be equal to 5 plus and then I take the antiderivative. The antiderivative of negative 2x is negative x squared. So we have negative x squared. And the antiderivative of 3 is just going to be 3x. So plus 3x. And we're going to evaluate it from 0 to 5/2. So this is going to be equal to 5 plus-- and I'll do all this stuff right over here. I'll do this stuff in green. So when we evaluate it at 5/2, this is going to be negative 5/2 squared. So it's going to be negative 25 over 4 plus 3 times 5/2, which is 15 over 2. And then from that, we're going to subtract this evaluated at 0. But negative 0 squared plus 3 times 0 is just 0. So this is what it simplifies to. And so what do we have right over here? So let's get our ourselves a common denominator. Looks like a common denominator right over here could be 4 So this is equal to-- 5 is the same thing as 20 over 4 minus 25 over 4 and then plus 30 over 4. So 20 plus 30 is 50 minus 25 is 25. So this is equal to 25 over 4. And 25 over 4 is the same thing as 6 and 1/4. So when we evaluate our function at this critical point, at this thing where the slope, or the derivative, is equal to 0, we got 6 and 1/4, which is higher than six, which is what g was at this end point. And it's definitely higher than what g was a negative 4. So the x-coordinate of the point at which g has an absolute maximum, on the interval negative 4 to three is x is equal to 5/2.