Absolute Maximum and Minimum in Calculus: Solving Problems with Graphs

[A_Munk3y]

Homework Statement

i did it on paint and print screened to make it more clear what i did :)

I am getting, as shown above, 0 for abs max and -15 for absolute min.
But it wants -48 for abs min and i have no idea how to get that >:(

 

[fzero]

The equation

$$f'(x) = \frac{x}{8} - \frac{8}{\sqrt{x}} = 0$$

has a solution in the relevant interval so there is a critical point.

 

[A_Munk3y]

ummm... can you explain that a little more lol
i don't really get what you mean :P

 

[fzero]

A critical point will be a local maximum or minimum. You need to compare the value of f at the critical point to the values of f on the boundary to find the absolute max/min.

 

[A_Munk3y]

16?
that would make the equation = 0
and if you plug it into g(x) you get 48

correct?

 

[fzero]

Is g(16)=48 or something else? Is x=16 a maximum or minimum?

 

[A_Munk3y]

x=16 is a minimum i meant g(x) = -48
and x = 0 is abs max b/c g(x) = 0
:D