Question:
Find the slope of the tangent to the curve $y=x^{3}-3 x+2$ at the point whose $x$-coordinate is $3 .$
Solution:
The given curve is $y=x^{3}-3 x+2$.
$\therefore \frac{d y}{d x}=3 x^{2}-3$
The slope of the tangent to a curve at $\left(x_{0}, y_{0}\right)$ is $\left.\frac{d y}{d x}\right\rfloor_{\left(x_{0}, y_{0}\right)}$.
Hence, the slope of the tangent at the point where the x-coordinate is 3 is given by,
$\left.\left.\frac{d y}{d x}\right]_{x=3}=3 x^{2}-3\right]_{x=3}=3(3)^{2}-3=27-3=24$