Question

answer the questions for the function f(x)= - 3x^3 + 3x^2 - x - 5

a. find formulas for f(x) and f(x)
f(x)= - 9x^2 + 6x - 1
f(x)= - 18x + 6
enter f(x), f(x), and f(x) into your grapher to examine the table.
b. set the first derivative f(x)=0 to find any critical values. tip: you may wish to graph f(x) by itself to explore if it has any zeros. it can be helpful to know that the formula of f(x) factors. select the correct choice below and, if necessary, fill in the answer boxes within your choice

a. there are no critical numbers

b. the function f has a critical number at x =, at this critical number, the second derivative f(x)=(type an exact answer.)

Explanation:

Step1: Factor the first - derivative

Given \(f'(x)=-9x^{2}+6x - 1\), we can rewrite it as \(f'(x)=-(9x^{2}-6x + 1)\). Using the perfect - square formula \((a - b)^2=a^{2}-2ab + b^{2}\) where \(a = 3x\) and \(b = 1\), we get \(f'(x)=-(3x - 1)^{2}\).

Step2: Find the critical values

Set \(f'(x)=0\), so \(-(3x - 1)^{2}=0\). Solving for \(x\), we have \((3x - 1)^{2}=0\), then \(3x-1 = 0\), and \(x=\frac{1}{3}\).

Step3: Find the second - derivative value at the critical point

Given \(f''(x)=-18x + 6\), substitute \(x = \frac{1}{3}\) into \(f''(x)\). Then \(f''(\frac{1}{3})=-18\times\frac{1}{3}+6=-6 + 6=0\).

Answer:

B. The function \(f\) has a critical number at \(x=\frac{1}{3}\), at this critical number, the second derivative \(f''(x)=0\)