Question

Find the values of x that are critical numbers of the function f(x)=3x⁴−16x³/6x²−72x+1

a) x=0,4,8
b) x=−1,0,6
c) x=1,4,8
d) x=−1,4,8

Answer 1

To find the critical numbers of the given function, we find the derivative of the function and set it equal to zero. Solving the resulting equation gives us the critical numbers of the function. The correct answer is not mentioned in the options.

Step-by-step explanation:

To find the values of x that are critical numbers of the function f(x)=3x⁴−16x³/6x²−72x+1, we need to determine the values of x for which the derivative of the function is equal to zero.

First, we find the derivative of the function:

f'(x) = 12x³ - 48x² - 72

Setting f'(x) = 0 and solving for x, we get:

12x³ - 48x² - 72 = 0

Dividing both sides of the equation by 12, we have:

x³ - 4x² - 6 = 0

Now we can solve this quadratic equation by factoring or using the quadratic formula. By factoring, we find:

(x - 2)(x + 1)(x - 3) = 0

Setting each factor equal to zero, we get the critical numbers:

x = 2, x = -1, x = 3