Question

Given f'(x)=(x-4)(4-2x), find the x-coordinate for the relative maximum on the graph of f(x)

Answer 1

f'(x)=0

Answer 2

How do you find the critical values? As Sparrow suggests, set the deriv. = to zero and solve for x.

How would you know whether you've located max, min or both?
I'd use the 2nd derivative. How does that work?

Answer 3

yeah but they are asking for f(x) so i think she have to first convert the function into f(x)

Answer 4

you can't convert there are infinite convertions

Answer 5

and you can't find f(x) only find x value

Answer 6

well then i guess if the curve is max it will become min and vice versa

Answer 7

oh i am sorry i though it was f inverse it actually is derivative of f(x)

Answer 8

x= 4,2

Answer 9

Again: Please set f '(x) (which you already have) = to 0 and solve for x. there will be 2 roots, which are y our critical values.

You certainly can integrate f '(x) to learn the exact shape of the graph of f(x), but since there is a constant of integration, that shape could be positioned anywhere along the y-axis.

Given f'(x)=(x-4)(4-2x)=0, find the 2 roots. Do that now, please.

Your next task will be to determine which of the 2 roots represents the x-coordinate of the relative minimum.

Answer 10

I won't check; I'll take your word for it that the critical values are 4 and 2.

Now that you know the critical values, how would you go about determining which one corresponds to a relative minimum in the function f(x)? I dropped you a hint earlier.

Answer 11

@chris215: Are you still interested in solving this problem?

Answer 12

Im not sure how to do that..?