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
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
Slope of the line $\displaystyle 3x-y=2$ is $\displaystyle 3$.
$\displaystyle y=x^3+2x+2$
$\displaystyle \frac{dy}{dx}=3x^2+2$
$\displaystyle \frac{dy}{dx}=3$
So, $\displaystyle 3x^2+2=3$
$\displaystyle 3x^2=1$
$\displaystyle x=\pm\frac{1}{\sqrt{3}}=\pm0.58$
The points on the curve: $\displaystyle y=x^3+2x+2$ whose tangent line is parallel to this line: $\displaystyle 3x-y=2$ are $\displaystyle (0.58, 3.35)$ and $\displaystyle (-0.58, 0.645)$.