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# Mean value theorem example: square root function

AP.CALC:
FUN‑1 (EU)
,
FUN‑1.B (LO)
,
FUN‑1.B.1 (EK)

## Video transcript

let f of X be equal to the square root of 4x minus 3 and let's see be the number that satisfies the mean value theorem for f on the closed interval between 1 and 3 or 1 is less than or equal to X is less than or equal to 3 what is C so let's just remind ourselves what it means for C to be the number that satisfies the mean value theorem for f this means that over this interval C is a point x equals C is a point where the slope of the tangent line at x equals C so I could write F prime of C so that is the slope of the tangent line when X is equal to C this is equal to the slope of the secant line that connects these two points so this is going to be equal to C the slope of the of the secant line that connects the points 3 f of 3 and 1 f of 1 so it's going to be f of 3 minus f of 1 over 3 minus 1 and if you wanted to think about what this means visually it would look something like this so this is our x-axis and this is 1 2 actually let me spread it out a little bit more 1 2 & 3 and so you have 1 comma F of 1 right over there so that is at the point 1 comma F of 1 and we could evaluate that actually what that's 1 comma 1 right so that's going to be the point 1 comma 1 and then you have the point 3 comma let's see you're going to have 4 times 3 is 12 minus 3 is 9 so it's going to be 3 comma 3 3 comma 3 so maybe it's right over there 3 comma 3 and it might look the curve might look something like this so it might look something like that so if you think about the slope of the line that connects these two points so this line that connects those two points all the mean value theorem a different color all the mean value theorem tells us is that there's a point between 1 and 3 where the slope of the tangent line has the exact same slope so if I were to eyeball it it looks like it's right around there although we are actually going to solve for it so some point where the slope of the tangent line is equal to the slope of the line that connects these two end points and their corresponding their corresponding function values so that is C that would be C right over there so really we just have to solve this so let's first just find out what f prime of X is and then we could substitute a C in there and then we can evaluate this on the right-hand side so I'm going to rewrite f of X f of X is equal to and I'm going to write it as 4 X to the minus 3 to the 1/2 power makes a little bit more obvious that we can apply the power rule and the chain rule here so f prime of X F prime of X is going to be the derivative of 4x minus 3 to the 1/2 with respect to 4x minus 3 so that is going to be 1/2 times 4x minus 3 to the negative 1/2 and then we're going to multiply that times the derivative of 4x minus 3 with respect to X well derivative of 4x with respect to X is just 4 and the derivative of negative 3 with respect to X well that's just going to be 0 so the derivative of 4x minus 3 with respect to X is 4 so times 4 so if F prime of X F prime of X is equal to 4 times 1/2 which is 2 over the square root of 4x minus 3 for X minus 3 to the 1/2 would just be the square root of 4x minus 3 but it's a negative 1/2 so we're going to put it in the denominator right over here and so f prime of C we could rewrite this as 2 over the square root of 4 C minus 3 and what is that going to be equal to that is going to be equal to let's see f of 3 we already figured out is three F of one we already figured out is one and so we get 3 minus 1 over 3 minus 1 well that's going to be 2 over 2 which is equal to 1 so there's some point between 1 & 3 where the derivative at that point the slope of the tangent line is equal to 1 so let's see if we can solve this thing right over here well we can multiply both sides of this by 4 C by the square root of 4 C minus 3 and so then we are going to get we're going to get 2 is equal to the square root of 4 C minus 3 all I did is multiply both sides of this by square root of 4 C minus 3 to get rid of this in the denominator and so let's see now to get rid of the radical we can square both sides and so actually let me just show that so now we can square both sides so we get 4 is equal to 4 c minus 3 add 3 to both sides 7 is equal to 4 C and then divide both sides by 4 I'll go right here to do it you're going to get c is equal to 7/4 c is equal to is equal to 7 over 4 which is equal to 1 and 3/4 or we could view this as 1.75 so actually the C value is a little bit closer I hand drew this it's closer to about right over there on our diagram and actually that looks pretty pretty close that actually looks pretty good I just hand drew this curve so it's definitely not exact but anyway hopefully that gives you a sense of what what's going on here we're just we're just saying the mean value theorem gives us some C where the slope of the tangent line is the same as the slope of the line that connects 3 1 f of 1 & 3 f of 3
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