Find the point on the curve y=x3−11x+5 at which the tangent is y=x-11
The equation of the curve is y=x3−11x+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
dydx=3x2−11
3x2−11=1 (Slope of given tangent =1)
⇒3x2−11=1⇒x2=4⇒x=±2
When x=2, then from Eq. (i), we get y=23−11×2+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)