Question

Find the point on the curve y = x³ − 11x + 5 at which the tangent is y = x − 11.

Answer 1

The equation of the given curve is y = x³ − 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)

∴ Slope of the tangent = 1

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

\(\frac{dy}{dx}\)=3x²-11

Then, we have:

3x²-11=1

3x²=12

x²=4

x=±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).