# Math Help - Intro Calculus - Curve Sketching

1. ## Intro Calculus - Curve Sketching

Sketch a graph of a function f(x) that is differentiable and that satisfies the following conditions.

a) $f'(x) > 0$, when $x < -3$ and $x > 1$

I don't know what to do... I'm used to sketching functions with equation questions and not interval questions....

Please show me a step by step solution to this question, thanks

2. Hello,

Originally Posted by Macleef
Sketch a graph of a function f(x) that is differentiable and that satisfies the following conditions.

a) $f'(x) > 0$, when $x < -3$ and $x > 1$

I don't know what to do... I'm used to sketching functions with equation questions and not interval questions....

Please show me a step by step solution to this question, thanks
This means that in the part of the sketch where x<-3, the function will be increasing (derivative is positive). It will be the same for x>1.

Between -3 and 1, the function is, for example, decreasing.

3. Originally Posted by Moo
Hello,

This means that in the part of the sketch where x<-3, the function will be increasing (derivative is positive). It will be the same for x>1.

Between -3 and 1, the function is, for example, decreasing.
Why is the section between -3 and 1 decreasing? Isn't it increasing since f'(x) is greater than 0?? I'm confused...

4. Originally Posted by Macleef
Why is the section between -3 and 1 decreasing? Isn't it increasing since f'(x) is greater than 0?? I'm confused...
You said it was increasing for x<-3, that is to say to the left of -3, and for x>1, that is to say to the right of 1.

Between -3 and 1, you can choose

5. so it doesn't matter? would the graph look like a positive cubic function?

6. Originally Posted by Macleef
so it doesn't matter? would the graph look like a positive cubic function?
yeah, whatever... You're only asked that it increases in the given intervals.
As the question is formulated, the function is just not increasing between -3 and 1 (so it can be constant or decreasing)

7. thanks for the help... and I read the question wrong... there's actually more to it than that...