Question

Given the function f(x)=2x3−12x2+18x, find the critical points.

A) x=0,3
B) x=1,6
C) x=2,4
D) x=−1,5

Answer 1

The critical points of the function f(x)=2x^3-12x^2+18x are found by setting its first derivative, f'(x)=6x^2-24x+18, equal to zero and solving the quadratic equation, which yields x=1 and x=3. These points are where the slope of the function is zero.

Step-by-step explanation:

To find the critical points of the function f(x)=2x^3−12x^2+18x, we need to find the values of x where the first derivative of the function is zero or undefined. To do this, we first take the derivative of f(x):

f'(x) = 6x^2 - 24x + 18

To find the critical points, we set the derivative equal to zero and solve for x:

0 = 6x^2 - 24x + 18

Dividing by 6 to simplify the equation, we get:

0 = x^2 - 4x + 3

This is a quadratic equation and we can factor it:

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

Setting each factor equal to zero gives us the critical points:

x = 3 and x = 1.

Therefore, the correct answer is x=1,3 which is not listed among the provided options A, B, C, D, implying a possible typo in the options or the question.