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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 the general 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 a function could look something like this and in this case it's absolute maximum is going to occur at the beginning of the interval or a function could look something like this it could look something like that and then the absolute maximum could occur at the endpoint of the interval or the other possibility is that the function looks something like the function looks something like 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 that the function is not differentiable there you could imagine a function that looks like this a function that looks like this and it maybe wouldn't be differentiable there but this still would be that still 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 what how high it gets or how large of a value we get for the G at the endpoints 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 at the different endpoints so let's start off let's evaluate G at negative 4 at the stand 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 two times negative 4 is negative 8 let me do it over here so I have some 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 let's change the bounds of integration here especially so that we can get the 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 this expression right here can be re-written as the negative of the integral between negative 4 and 0 F of T DT and now this this expression right over here is the area under F of T or in this case f of X is or the area under F between negative 4 & 0 so it's this 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 circle the area of this circle 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 would be 9 pi 9 pi and of course we're going to divide it by 4 multiplied 1 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 the area of this entire circle we have a radius of 1 so it's going to be pi times 1 squared but just PI R squared so it's going to be PI and we're gonna divide it by 4 we're going to divide it by 4 because 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 we're it's a quarter circle in either case and we are subtracting it because the area is below the 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 I do that right 1/4 times 8 pi so this all simplifies to 2 pi so G of negative 4 G of negative 4 is equal to negative 8 minus negative 8 minus 2 pi so clearly a negative number here more negative than negative 8 so let's try the other bounds let's see what G of positive 3 is let's see what G of positive 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 can have this positive area and then it's going to become 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 we already found the end point where G hits a positive value so negative 4 is definitely not a candidate 3 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 any critical points that G has in between so points that are it's either undifferentiated is equal to 0 so let's look at it if so G prime G prime of X we just take the derivative of this business up here derivative of 2x is 2 derivative of this definite integral 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 this is just going to be this is just going to be plus 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 you give any x value over this interval we have a value for f of X f of X isn't differentiable everywhere but there's a 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 at the interval so G is actually differentiable throughout the interval so the only critical points would be where this derivative is equal to zero so let's set this thing equal to zero 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 zero or two plus f of X is equal to zero you can subtract two from both sides and you get f of X is equal to negative two so any X that satisfies f of X is equal to negative two is a point where this where the derivative of G is equal to zero and let's see if if f of X is equal to negative two at any point so let me draw a line over here negative two and negative two we have to look at it visually because they've only given us this visual definition of f of X doesn't equal negative two Justin equal to negative two only equals negative two right right over there right over there and it looks like we're at about two and a half 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 two and we could figure out the slope of this line fairly visually we can figure or figure out the equation of this line fairly visually we can figure out slope if we run if we run three so change in X if our change in X is three then our change in Y our rise rise is negative six change in Y is equal to negative six slope is rise over run or change in Y over change in X so negative six divided by 3 is negative two it has a slope of negative two and that actually I could have done that easier where if we go forward one we go down by two so we could have seen that the slope is negative two so this this part of f of X we have Y is equal to negative two x plus and then the y-intercept is pretty straightforward this is at 3 1 2 3 plus negative 2 X plus 3 so where where the 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 negative 2x plus 3 is equal to 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 negative 2x plus 3 would have to have some some form of equation for these circles but over it right over here this is what f of X is and now we can solve this pretty pretty straightforwardly so we could subtract 3 from both sides we could 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 halves which is exactly what we thought it was when we looked at it when we looked at it visually it looked like we were at about 2 and 1/2 which is the same thing as 5 halves now we don't know 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 let's evaluate G at 5 halves so G at 5 halves is going to be equal to 2 times 5 halves 2 times 5 halves plus plus the integral from 0 to 5 halves from 0 to 5 halves of F of T F of T DT so this first part right over here the twos cancel out the twos cancel out so this is going to be equal to this is going to be equal to 5 and then plus the integral from 0 to 5 halves 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 the the equation for it over this interval so it's the integral of negative 2x plus 3 tea and then let's just evaluate this let me get some real estate so this is me draw a line here so we don't get confused so this is going to be equal to five plus and then take the antiderivative the antiderivative of negative two x is negative x squared so negative let me do it 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 halves so this is going to be equal to 5 plus and I'll do all of this stuff right over here I'll do this stuff in green so when we evaluate it at 5 halves this is going to be negative of 5 1/2 squared so it's going to be negative 25 over 4 plus 3 times 5 halves which is 15 over 2 and then we're from that we're going to subtract this evaluated at 0 but negative 0 squared plus 3 times 0 is just is just a 0 so this is what it simplifies to and so what do we have what do we have right over here so let's get ourselves let's get ourselves a common denominator let's click a common denominator right over here it 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 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 were the slope where the derivative is equal to zero we got six and one fourths which is higher than 6 which is what G was at this endpoint and it's definitely higher than what G was at negative 4 so the x-coordinate the x-coordinate of the point at which G has an absolute maximum on the interval negative 4 to 3 is it is X is equal to 5 halves
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