Question

PLEASE HELP ME WITH CALCULUS!!!!

Answer 1

the question is....?

Answer 2

MAX and MIn-

f(x) = x^3/ x + 2 I= [-1,1]

Answer 3

It's something like this, you need to derivate it, when the derivative equals zero, thats a maximus or minimum value of the func.

So d/dx(f) = ((x+2)3x^2 + x^3)/(x+2)^2

That, equals zero, solve and check what do you get.

Answer 4

\[f(x)=\frac{x^3}{x+2}\]
\[f'(x)=\frac{2x^2(x+3)}{(x+2)^2}\] zeros are 0 and -3, but only 0 is in your domain

Answer 5

If the second derivative, at that point, is negative, then it's a maxima

Answer 6

Oh right, somethink like that

Answer 7

Uhmm, look to the right and then to the left, is an inflection point

Answer 8

you don't need the second derivative for this problem
you are asked for a max and min on a closed interval.
check
\[f(-1),f(0),f(1)\] mas is the max, min is the min, that is all