# Math Help - Little challenge for persons who know Calculus and Geometry

1. ## Little challenge for persons who know Calculus and Geometry

I've found hard to answer this problem, I would apreciate your help:

Find the points on the graph of f(x) = 1/3x^3+x^2-x-1 at which the slope is (a) -1, (b) 2, and (c) 0

2. Originally Posted by yasu7
I've found hard to answer this problem, I would apreciate your help:

Find the points on the graph of f(x) = 1/3x^3+x^2-x-1 at which the slope is (a) -1, (b) 2, and (c) 0
Find $f'(x)$:

$f'(x)=x^2+2x-1$

Now find $x$ when $f'(x)=-1$, $f'(x)=2$ and $f'(x)=0$

--Chris

3. for example for $f'(x) = 0$ the solution is like this:

$x^2+2x-1=0$
$X^2+2x+1-2=0$
$(x+1)^2-2=0$
$(x+1)^2=2$
$x = -1+\sqrt2$
$x= -1-\sqrt2$