Question

Use the graph of f '(x) below to find the x values of the relative maximum on the graph of f(x):

Answer 1

is it 2?

Answer 2

is there supposed to be another graph f(x)

Answer 3

noo i dont think so?

Answer 4

The way the problem reads has confused me.

Answer 5

It appears to me it asking for a point on the graph of f(x) based on the graph of f'(x)??

Answer 6

Help us @Astrophysics

Answer 7

I would think we have to consider the second derivative test

Answer 8

ok idk how to do that

Answer 9

Well the derivative here will be the slope, if we consider as such we may think of the second derivative, which says,
if the slope is zero at x, the second derivative at x:
If <0 -> local max
>0 - > local min
=0 -> test fails

Answer 10

I understand that min or max can occur when F'(x) = 0
Here it equals 0 when x = 0, x=1, and x = 2

Answer 11

One of those (or maybe 2) will probably be where the maximum would occur.

Answer 12

Yes, that's what gets me

Answer 13

But it is a sine wave and F"(x) would be cos x

Answer 14

Right

Answer 15

ok bc im confusedd

Answer 16

I would think if we consider it as sinx then it would be clear, but I'm not sure, been too long since I've done calc 1 haha.

Answer 17

Well from 0 to 90 cos x would go from 1 to 0
from 90 to180 cos x would go to 0 to -1
from 180 to 270 cos x goes from -1 to 0
from 270 to 360 cos x goes from 0 to 1

So the maximum positive is at 0, and 360
Now am I to conclude that the max occurs at 180 degrees which would be point x =1 on the graph, still confused and pondering.

Answer 18

That would be the point where Cos x is negative with a -1...........

Answer 19

hmm ok thanks for helping :)

Answer 20

Sorry chris that I am not to confident with this, it has been a long time ago.
Good luck with your studies.

Answer 21

Thats alright I'll just ask my teacher :)

Answer 22

Yes, that is what they are for to help.

Answer 23

that graph of \(f'(x)\) looks like a graph of \(y = 2\sin \pi x\)

but \(2 \sin \pi x\) is the derivative of \[y=-\frac{2}{\pi}\cos \pi x\]which has a maximum of \[y=\frac{2}{\pi}\] at \(x=1\) where \(\cos \pi x = -1\)

we know that at a maximum or minimum, the derivative is equal to 0. If you think of a ball thrown up in the air, the derivative of the height is positive while the ball travels upward, becomes 0 at the moment the ball is at the apex of he flight, and then becomes negative as the ball starts down to the ground.

Answer 24

because the derivative changes from being positive to negative at x=1, that means we have changed from going up to going down, and the point in between was a local maximum. if you look at the graph at x = 2, there you see that the graph of the derivative goes from negative to positive, and indeed, x = 2 is a point where \[y=-\frac{2}{\pi}\cos \pi x\]has a minimum.

I encourage you to plot that function and see for yourself!

Answer 25

http://www.wolframalpha.com/input/?i=plot%20%202%20sin%20%CF%80x%2c%20-2%2f%CF%80%20cos%20%CF%80x%20from%20x%20%3d0%20to%20x%20%3d2.2

if you are only willing to click on a link