Question

Find the maximal and minimal values of the function

Answer 1

find the derivative of that using the definition of derivative, set it equal to 0, then solve for x, those are your local extremas. remember to test them...

Answer 2

f(1/32)= 96.016
f(2) = 5.743
f(32) = 64.375

how do i tell if its maximum or minimum?

Answer 3

like i side find f'(x) not f(x)

Answer 4

said*

Answer 5

f'(x) = 6/5x^1/5 - 24x^-9/5

Answer 6

*24/5

Answer 7

i found x = 2. but how to test if those values are max or min?

Answer 8

plug in 1 and 3 for x, if u it's - for 1 and + for 3 then it's minimum, if u get + for 1 and - for 3 it's maximum

Answer 9

+ being positive and - negative

Answer 10

@11calcBC why the choice of numbers? should i plug it in to f(x) or f'(x)?

Answer 11

it can be any number to the left and right of 2, plug them into f'(x)

Answer 12

if the result is -, that means the slope at that point til 2 is negative ()going down), vice versa, u get a minimum becuz the graph is going down til 2, then gooing back up after 2. the other way mean's a max

Answer 13

the slope is 0 at only 1 point in the interval [ 1/32, 32]
that means the function has 1 max (or 1 min) in this interval.
if the value f(x) at 1/32 is bigger than f(2), then f(2) is the MIN

to find the max, you have to look at the boundaries, one or the other will be the max in the given interval. As you know, f(1/32) is bigger than f(32), so f(1/32)= 96.016 is the MAX

Answer 14

ah thanks guys it clear now(: i tried both methods, but @phi is more straightforward. thanks!(:

Answer 15

lol yea oka as long as u get it

Answer 16

you could use the 2nd derivative to decide if f(2) is a min or max, but because you also have to evaluate the boundary values , it is an unneeded complication

Answer 17

yeah i tried to calculate for 2nd derivative also, but i think its very easy to mess up the calculations due to the powers. so i was finding for a better way :p thanks!