Question

If f(x)=6x²-x³, then at what value(s) of x does the tangent to the curve f(x)is parallel to line 9x=1+y?

Answer 1

The given line in slope-intercept form is

9x = 1 + y   ⇒   y = 9x - 1

so that its slope is 9. Any tangent to f(x) at some point (x, y) will be parallel to this line whenever f'(x) = 9 at that point.

We have

f(x) = 6x² - x³

with derivative

f'(x) = 12x - 3x²

Solve for x that makes f'(x) = 9 :

9 = 12x - 3x²

3x² - 12x + 9 = 0

x² - 4x + 3 = 0

(x - 3) (x - 1) = 0

x = 3 or x = 1