Question

Find the point on the curve $y=x^3-11x+5$ at which the tangent is $y=x-11.$

Answer 1

The equation of the given curve is $y=x^3-11x+5$.

The equation of the tangent to the given curve is given as $y=x-11$
(which is of the form $y=mx+c$).

$\therefore$ Slope of the tangent $=1$

Now, the slope of the tangent to the given curve at the point $(x,y)$ is given by,

$\frac{dy}{dx}=3x^2-11$

Then, we have:

$3x^2-11=1\Rightarrow 3x^2=12$

$\Rightarrow x^2=4$

$\Rightarrow x=\pm 2$

When $x=2,y=(2{)}^3-11(2)+5=8-22+5=-9$.

When $x=-2,y=(-2{)}^3-11(-2)+5=-8+22+5=19$.

Hence, the required points are $(2,9)$ and $(2,19).$