Question

the function ( f(x)=2 x^{3}-33 x^{2}+60 x - 8 ) has two critical numbers. the smaller one is ( x = ) and the larger one is ( x = )

Explanation:

Step1: Find the derivative of the function

The derivative of \(f(x)=2x^{3}-33x^{2}+60x - 8\) is \(f^\prime(x)=6x^{2}-66x + 60\).

Step2: Set the derivative equal to zero

Set \(f^\prime(x)=0\), so \(6x^{2}-66x + 60 = 0\). Divide through by \(6\) to get \(x^{2}-11x + 10=0\).

Step3: Factor the quadratic equation

Factor \(x^{2}-11x + 10=(x - 1)(x - 10)=0\).

Step4: Solve for \(x\)

Using the zero - product property, if \((x - 1)(x - 10)=0\), then \(x=1\) or \(x = 10\).

Answer:

The smaller one is \(x = 1\) and the larger one is \(x=10\).