Question

Let f be the function with the first derivative defined by f'(x) = sin(x^3) for 0≤x≤2.
determine the value of x for which f attains its absolute maximum value on the closed interval 0≤x≤2? Explain why this is the absolute maximum.

Answer 1

We know the max, min, inflection points occur when f' = 0 or undefined.

Answer 2

f' > 0 then f is increasing. f' < 0 then f is decreasing. Think of a maximum like a mountain, first you go up, then you go down. So we expect f' to go from a positive to a negative value at a max point.

Answer 3

so there is a f(x) max at x = 1.465 ?

Answer 4

but how would I figure out if it's the absolute max? :3

Answer 5

First consider y = x^3, which ranges as 0^3≤x^3≤2^3 => 0≤y≤8
Where would the roots be of sin(y)?

Answer 6

You should have y = 0, pi, 2pi. This means x = 0, (pi)^(1/3), (2pi)^(1/3)

Answer 7

We need to determine which ones are actual candidates for the maximum.

Answer 8

erm
so end points and critical points of f(x)
so 0 ---- 1.465 ---- 1.845 ---- 2

Answer 9

then do the derivative test thing?
so f(x) increasing then decreasing then increasing?

Answer 10

so only 1.465 and 2 could possibly be maxes

Answer 11

increase -> decrease means there is a max

Answer 12

so x = 1.465 is the absolute max on interval 0≤x≤2 ?

Answer 13

Well, you don't need to go to a decrease state for an endpoint technically. For the start point you would be in a decrease state, for the final point you would be in an increase state. That would make you candidates for max

Answer 14

yeah so 1.465 and 2 can be maxes

Answer 15

Yes, and I don't know how you would single one out without integration to be honest.

Answer 16

yeah, my teacher hadn't taught integrals yet, but he said that the area under the curve shows the amount of increase or decrease or whatever.

so since the area of increase before 2 is less than the area of decrease after 1.465
therefore 1.465 is the absolute max between 0 and 2 ?

Answer 17

well int a->b of f' = f(b) - f(a). If f(b)-f(a)>0, then f(b)>f(a)

Answer 18

erm.... ...
so did I get it right?
since the area of increase before 2 is less than the area of decrease after 1.465
therefore 1.465 is the absolute max between 0 and 2 ?

Answer 19

dunno what you mean by area of decrease

Answer 20

like the area under the curve stuff

Answer 21

look at area under the curve between 1.465 and 2

Answer 22

If increasing, then 2 is bigger, if decreasing 1.465 is bigger

Answer 23

so it's 1.465?
http://www.wolframalpha.com/input/?i=sin%28x%5E3%29+%3D+0++between+0+and+2

Answer 24

Between 0 and 1.465, the area is positive. --- (1)
Between 1.465 and 1.845, the area is negative. ---- (2)
Between 1.845 and 2, the area is positive. ----- (3)
But more area is subtracted in (2) than is added in (3) and so there is a net loss in area by the time we get to x = 2.

So f(x) must attain absolute max at x = 1.465

Answer 25

ok, thank you for your help! @ranga @wio