the function $f(x)=2x^{3}-27x^{2}+108x - 8$ has two critical values. the smaller one equals and the larger…

the function $f(x)=2x^{3}-27x^{2}+108x - 8$ has two critical values. the smaller one equals and the larger one equals

the function $f(x)=2x^{3}-27x^{2}+108x - 8$ has two critical values. the smaller one equals and the larger one equals

Answer

Explanation:

Step1: Find the derivative

Differentiate $f(x)=2x^{3}-27x^{2}+108x - 8$ using the power - rule $(x^n)'=nx^{n - 1}$. $f'(x)=6x^{2}-54x + 108$

Step2: Set the derivative equal to zero

$6x^{2}-54x + 108 = 0$. Divide through by 6: $x^{2}-9x + 18=0$

Step3: Solve the quadratic equation

Factor the quadratic equation: $x^{2}-9x + 18=(x - 3)(x - 6)=0$. Using the zero - product property, if $(x - 3)(x - 6)=0$, then $x-3 = 0$ or $x - 6=0$. So $x = 3$ or $x = 6$

Answer:

The smaller one equals 3 and the larger one equals 6