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 curve is $y=x^3-11x+5$ ....(i)

The equation of the tangent to the given curve is y=mx+c(slope intercept form)

Given line is y=x-11

On comparing, we get slope of the tangent, m=1

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

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

$3x^2-11=1$ (Slope of given tangent =1)

$\Rightarrow 3x^2-11=1\Rightarrow x^2=4\Rightarrow x=\pm 2$

When x=2, then from Eq. (i), we get $y=2^3-11\times 2+5=-9$

When x=-2, then from Eq. (i), we get y=(-2)^3-11(-2)+5=19

Equation of tangent at point (2,-9) is y-(-9)=1(x-2) or y=x-11.

But the point (-2,19) does not lie on the line y=x-11.

The answer is (2,-9)