Slope of tangent line

**ascendancy523** · June 8th 2010, 11:15 AM

I was wondering if someone could check my work. Here's my problem:

Find the slopes of the tangent line to the curve $y=x^3-3x$ at the points where x=-2,-1,0,1,2.

So $m_{tan}=\frac{[(c+h)^3-3(c+h)]-(c^3-3c)}{h}$

$=\frac{(c^3+3c^2h+3ch^2+h^3)-3c-3h-c^3+3c)}{h}$

$=\frac{3c^2h+3ch^2-3h+h^3}{h}$

$=\frac{h(3c^2+3ch-3+h^2)}{h}$

lim h->0 $3c^2+3ch-3+h^2=3c^2-3$

so at x=-2, $m_{tan}=9$
x=-1, $m_{tan}=0$
x=0, $m_{tan}=-3$
x=1, $m_{tan}=0$
x=2, $m_{tan}=9$

Thank you for any help!

**Ackbeet** · June 8th 2010, 11:17 AM

Looks good to me.

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