1. ## Finding points...

Find points on the graph of the function $\displaystyle f (x)= x^3-2$ where the slope is 3.

My first instinct would be to do the derivative of $\displaystyle f (x)= x^3-2$, which would be $\displaystyle 3x^2$. But this does not help in finding the points of the function where the slope is 3.

2. Set the derivative equal to the slope.

$\displaystyle F'(x)=3x^2=3$

x=1,-1

plug these 2 points back into the original equation to find the points.

3. So the points are $\displaystyle (1,-1) (-1,-3)$? I would never have figured that out.

4. Originally Posted by yeloc
I would never have figured that out.
why not? you want the slope to be equal to 3, the derivative gives the formula for the slope, so you want the derivative to equal 3.

if you know what the derivative is, it would not be that large a leap in logic i would hope

5. Originally Posted by yeloc
So the points are $\displaystyle (1,-1) (-1,-3)$? I would never have figured that out.
Yup

6. Originally Posted by Jhevon
why not? you want the slope to be equal to 3, the derivative gives the formula for the slope, so you want the derivative to equal 3.

if you know what the derivative is, it would not be that large a leap in logic i would hope
Hindsight is 20/20. Of course it seems obvious now.

7. Originally Posted by yeloc
Hindsight is 20/20.
haha, i like that saying. it sounds like it is a famous saying, or if not, it should be. i wonder why i've never heard of it before?

8. Originally Posted by Jhevon
haha, i like that saying. it sounds like it is a famous saying, or if not, it should be. i wonder why i've never heard of it before?
It's a cliched saying, so I'm sure you've heard it before. Thank you Jhevon and Linnus for your help.