 In this video, we provide the solution to question number 13 for practice exam number four for math 1210 We're given a function f of x which is the polynomial negative 4x cube plus 7x minus 3 We're supposed to first explain why f satisfies the hypotheses of the mean value theorem on the interval 0 1 So let's just answer that question first so the mean value theorem Expects that on your closed interval you're continuous on the closed interval and Differentiable on the open interval. So something like the following would be appropriate since f is Differentiable It's differential on its domain Which what is that domain the domain is negative infinity to infinity? F is continuous on The interval 0 to 1 this is a closed interval and it's going to be differentiable on The open interval 0 to 1 so first of all if it's differentiable everywhere Then it'll be differentiable on some smaller interval of course, but differentially also implies continuity So if it's differentiable everywhere, it's actually continuous everywhere and therefore it's continuous on the closed interval 0 to 1 so that answers the first thing so we should we shouldn't actually state that we've answered the first thing So something like so f satisfies the hypotheses The mean value theorem All right So then what's the next thing it says find any values? So find any values of C that satisfy the conclusion of the mean value theorem for f of x on the interval 0 to 1 If no such values exist explain why not well We've already said it satisfies the condition so therefore such a point has to exist Well, what points are guaranteed by the mean value theorem? So the mean value theorem is going to tell you that the derivative f prime at C is going to somewhere equal the average rate of change Delta y over delta x that is dy over dx equals delta y over delta x for some value here Which what is the average rate change in this situation? We're going to get f of 1 minus f of 0 over 1 minus 0 let's continue to compute this thing if I plug 1 into my function I'm going to get negative 4 plus 7 minus 3 and I'm going to subtract from that f of 0 Which is going to give you 0 plus 0 minus 3 all over 1 minus 0, which is just 1 So notice that the negative 3 will cancel out negative 3 minus negative 3 we cancel out those are just 0 so it's cancel out so negative 4 Plus 7 is going to be 3 over 1 so I get 3 so I get that somewhere the derivative is equal to 3 But if I compute the derivative for the function, I'm going to get negative 12 C squared plus 7 so I have to solve for this equation right here Which if I subtract 7 from both sides you get negative 12 C squared equals negative 4 divide both sides by 12 You get C squared equals 1 over 3 4 goes into 12 3 times of course and then taking the square root we get that C is Equal to plus or minus 1 over the square root of 3 now 1 over the square root of 3 Which is positive will be in the interval 0 to 1 but the negative is not that's be outside the interval so we can then conclude something like sense 1 over the square root of 3 is less than 1 but greater than 0 we get that We have that C is gonna be 1 over the square root of 3 This was the value guaranteed by the mean value theorem and this therefore answers the question presented