Question

Find the critical points of the function $f\left(x\right)=2x^3+15x^2+36x.$

• A. x = 2 , x = 3
• B. x = - 2 , x = 3
• C. x = 2 , x = - 3
• D. x = - 2 , x = - 3

Answer 1

The correct option is D

x = - 2 , x = - 3

Critical points are the points where f'(x) is zero or it doesn't exist. Since the given function is a polynomial function, we can say that it would be continuous and differentiable at all the points.

So, to find the critical points, we find all the zeroes of f’(x)

f’(x) = 6$x^2$+ 30x + 36

Or f’(x) = 6 ($x^2$+ 5x + 6)

Or f’(x) = 6 (x + 2) (x +3)

To find the critical points, f’(x) = 0

So, 6 (x + 2) (x +3 ) = 0

x = -2 , x = -3 are the points where f’(x) =0.

These are the only critical points of this function. The graph of function will look like this -