answer the questions for the function f(x)= - 3x^3 + 3x^2 - x - 5\n\na. find formulas for f(x) and…
answer the questions for the function f(x)= - 3x^3 + 3x^2 - x - 5\n\na. find formulas for f(x) and f(x)\nf(x)= - 9x^2 + 6x - 1\nf(x)= - 18x + 6\nenter f(x), f(x), and f(x) into your grapher to examine the table.\nb. 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\n\na. there are no critical numbers\n\nb. the function f has a critical number at x =, at this critical number, the second derivative f(x)=(type an exact answer.)
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)