1. ## Differentiation and Geometry

Hi, I ve been trying to solve this question for days, but my answer is not the same as in the book; could someone help, plz?!!

Q: Find the point(s) on the curve: y=x^3+2x+2 whose tangent line is parallel to this line: 3x-y=2

Thanksssss

2. Slope of the line $3x-y=2$ is $3$.

$y=x^3+2x+2$

$\frac{dy}{dx}=3x^2+2$

$\frac{dy}{dx}=3$

So, $3x^2+2=3$

$3x^2=1$

$x=\pm\frac{1}{\sqrt{3}}=\pm0.58$

The points on the curve: $y=x^3+2x+2$ whose tangent line is parallel to this line: $3x-y=2$ are $(0.58, 3.35)$ and $(-0.58, 0.645)$.

3. based on this, the points will be where x= +/- (1/sqrt 3)??!!

based on this, the points will be where x= +/- (1/sqrt 3)??!!
Yes.

5. ## Thankz

This means the text book is wrong then.
Tnx very very much..

This means the text book is wrong then.
Tnx very very much..
What does the textbook say?

7. It says:

x= +/- (1/sqrt2)