Question

Find the slope of the tangent to the curve $y=x^3-3x+2$ at the point whose $x$ coordinate is $3$.

Answer 1

The given curve is $y=x^3-3x+2$
$\Rightarrow \frac{dy}{dx}=3x^2-3$
Thus slope of the tangent to the given curve at $x=3$ is
${\left(\frac{dy}{dx}\right)}_{x=3}={(3x^2-3)}_{x=3}=3(3{)}^2-3=24$