1. ## Check my work?

Here's the problem:

"Find the x-coordinate of all points on the graph of $\displaystyle f(x)=-2x^3+2x+1$ where the tangent line is parallel to the line $\displaystyle y=-7x+2$."

I took the derivative of f(x) and got $\displaystyle f'(x)=-6x^2+2$. Then I set the slope of the line y equal to this equation, so as to have $\displaystyle -7=-6x^2+2$. So I solved for x and got $\displaystyle x^2=-(9/6)$ and so x=1.225 (after taking the square root).

I went ahead and found the actual tangent line of this by plugging that x-value back into the derivative of f(x), ending with $\displaystyle y=-7(x+1.225)+7$.

I'm pretty sure I did the process right, but if I didn't, could someone kindly correct me? Or tell me if I'm leaving something out? I feel like there's something else that should be done.

2. Originally Posted by Rumor
Here's the problem:

"Find the x-coordinate of all points on the graph of $\displaystyle f(x)=-2x^3+2x+1$ where the tangent line is parallel to the line $\displaystyle y=-7x+2$."

I took the derivative of f(x) and got $\displaystyle f'(x)=-6x^2+2$. Then I set the slope of the line y equal to this equation, so as to have $\displaystyle -7=-6x^2+2$. So I solved for x and got $\displaystyle x^2=-(9/6)$ and so x=1.225 (after taking the square root).

I went ahead and found the actual tangent line of this by plugging that x-value back into the derivative of f(x), ending with $\displaystyle y=-7(x+1.225)+7$.

I'm pretty sure I did the process right, but if I didn't, could someone kindly correct me? Or tell me if I'm leaving something out? I feel like there's something else that should be done.

$\displaystyle f'(x)=-6x^2+2=-7$

So

$\displaystyle -6x^2=-9$

$\displaystyle x^2=\frac{9}{6}$

Now you need both positive and negative square root

3. Originally Posted by Rumor
Here's the problem:

"Find the x-coordinate of all points on the graph of $\displaystyle f(x)=-2x^3+2x+1$ where the tangent line is parallel to the line $\displaystyle y=-7x+2$."

I took the derivative of f(x) and got $\displaystyle f'(x)=-6x^2+2$. Then I set the slope of the line y equal to this equation, so as to have $\displaystyle -7=-6x^2+2$. So I solved for x and got $\displaystyle x^2=-(9/6)$ {-9/6 ???} and so x=1.225 (after taking the square root).

I went ahead and found the actual tangent line of this by plugging that x-value back into the derivative of f(x), ending with $\displaystyle y=-7(x+1.225)+7$.

I'm pretty sure I did the process right, but if I didn't, could someone kindly correct me? Or tell me if I'm leaving something out? I feel like there's something else that should be done.
$\displaystyle -6x^2+2 = -7$

$\displaystyle x^2 = \frac{3}{2}$

$\displaystyle x = \pm \sqrt{\frac{3}{2}}$

you forgot the negative solution for x.

4. Oh, of course! I knew there was something.. Thanks to the both of you. :]