# Thread: 3rd degree equation and derivative

1. ## 3rd degree equation and derivative

$\displaystyle f(x)=0.002x^3-0.08x^2+x$
Interval: 0 < x < 50

I need to find the x-intervals, where the slope-angles are between 0 and 45 degrees with the x-axis' positive direction.

I'm kinda lost?

2. Originally Posted by No Logic Sense
$\displaystyle f(x)=0.002x^3-0.08x^2+x$
Interval: 0 < x < 50

I need to find the x-intervals, where the slope-angles are between 0 and 45 degrees with the x-axis' positive direction.

I'm kinda lost?
1. Draw a graph of y = f(x)

2. Find f'(x).

3. Draw a graph of y = f'(x).

4. Note that m = f'(x) = tan (theta) where theta is the angle between the x-axis's positive direction.

5. Solve tan(0) < f'(x) < tan(45) <=> 0 < f'(x) < 1 (all angle sin degrees). The graph of y = f'(x) might help to see what's happening with this inequation.

3. Originally Posted by mr fantastic
1. Draw a graph of y = f(x)

2. Find f'(x).

3. Draw a graph of y = f'(x).

4. Note that m = f'(x) = tan (theta) where theta is the angle between the x-axis's positive direction.

5. Solve tan(0) < f'(x) < tan(45) <=> 0 < f'(x) < 1 (all angle sin degrees). The graph of y = f'(x) might help to see what's happening with this inequation.
I'm not sure I undertand number 4. Can you elaborate?

4. Originally Posted by No Logic Sense
I'm not sure I undertand number 4. Can you elaborate?
m is the gradient of the tangent to the curve. It's given by f'(x).

The tangent is a line. Have you met the formula m = tan theta, where theta is the angle between a line and the x-axis's positive direction?

5. Alright, I understand now. Thank you.