Question

Find the absolute extrema of the function on the closed interval.
h(s)= 2/(s-4) [0, 2]
my work so far
f'(x)=(s-4)(0)/(s-4)^2 = -2/(s-4)^2
f(0)=0
f(0)=0/0-4 = 0
f(2)=2/2-4 =2/-2 =-1
at max (0,0) (-1, 2)
min (0,0)
but it was all wrong

Answer 1

Ok so you started by trying to find f'
...and f' is looks good \[f'(x)=\frac{(2)'(s-4)-(s-4)'(2)}{(s-4)^2} = \frac{-2}{(s-4)^2} \]

Answer 2

So the only two numbers to check is 0 and 2

Answer 3

did you plug 0 and 2 into the original function?

Answer 4

\[f(0)=\frac{2}{0-4}=? \text{ and } f(2)=\frac{2}{2-4}=?\]

Answer 5

f(0) doesn't equal 0

Answer 6

but f(2) does equal 2/-2=-1 so you did good there

Answer 7

f(0) doesn't equal 0 for sure because 2 will never be 0

Answer 8

\[f(0)=\frac{2}{0-4}=\frac{2}{-4}=\frac{-1}{2}\]

Answer 9

so you have the two points as absolute extrema (0,-1/2) and (2,-1)
One of them is the absolute max and the other is the abs min
hint: the one with the highest y-value is absolute max and the one with the lowest y value is the absolute min

Answer 10

So your work wasn't all bad

Answer 11

ok thank you

Answer 12

Do you have any questions?

Answer 13

no i figured it out thank you