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

Solution: Equation of the tangent is ,
$y=x-11$
Comparing it with $y=mx+c$,
$m=1$
So, slope of the tangent is $1$.
Now, equation of the curve is,
$y=x^3-11x+5$
$\therefore 1=\frac{dy}{dx}=3x^2-11$
$\Rightarrow 3x^2-11=1$
$\Rightarrow 3x^2=12$
$\Rightarrow x^2=4$
$\Rightarrow x=\pm 2$
When $x=2,y={\left(2\right)}^3-11\left(2\right)+5=-9$
If we put $x=2,y=-9$, it satisfy the equation, $y=x-11$. So, it is the required point.
When $x=-2,y={\left(-2\right)}^3-11\left(-2\right)+5=19$
If we put $x=-2,y=19$, it does not satisfy the equation, $y=x-11$.
So, the required point on the curve is $\left(2,-9\right)$.