Question

find the max and min value of
f(x) = x - [(8x)/(x+2)] on the interval [0,4]

Answer 1

Please help!!!

Answer 2

Have you considered the 1st derivative?

Answer 3

i did but im not sure what to do after that

Answer 4

So? Let's see the derivative. It is not necessary to see the very end. It is sufficient to know one step of the trip. You didn't do the 1st derivative the hardest way possible, did you?

Answer 5

i dont think so this is what i got
-8/(x+2) + (8x)/(x+2)^2 + 1

Answer 6

Find the critical values first and check if any of them lie in the given closed interval. If any of them do, find the value of f(x) at those critical values. If there is no critical values in the given interval then you don't need to care about them. After that evaluate the function at the endpoints of the given interval. Now compare the values of f(x) at the critical values (only if the critical values are in the given interval) with the values of the function at the endpoints of the interval. Whichever is the biggest is the maximum and the lowest is the minimum.

@MG757983

Answer 7

Well, okay, that IS the right derivative, but WOW!!!! Make your life a little simpler.

Use long division or partial fractions or something to rewrite the original expression.

\(f(x) = x - [(8x)/(x+2)] = x - 8 + \dfrac{16}{x+2}\)

Then, it is SO MUCH simpler to find the 1st derivative: \(f'(x) = 1 - \dfrac{16}{(x+2)^{2}}\).

Like I said, you DID get that final result. It's not real obvious, but it's there.

The point of the question is to find were the derivative is zero in the interval.

Solve: \(f'(x) = 0\)

Answer 8

There is a vertical asymptote at x = -2, so that should not bother you. f(x) is rather well-behaved on [0,4].

Answer 9

ahhhh okay i see what you guys are saying...the minimum value is going to be -2 and im working on the maximum value now

Answer 10

and that max value is 0.....THANK YOU SOOO MUUUCHHH

Answer 11

Ah, you seem to have investigated the endpoints! Very good work!