Question

answer the questions for the function f(x)= - 3x^3 + 3x^2 - x - 5
a. there are no critical numbers.
b. the function f has a critical number at x = 1/3; at this critical number, the second derivative f(x)=0 (type an exact answer.)
c. what conclusion can be made from your answer to part b? select the correct choice below and, if necessary, fill in the answer box within your choice
a. at the critical number x = the function f is concave up and corresponds to a relative maximum of f(x)
b. at the critical number x = the function f is concave down and corresponds to a relative maximum of f(x)
c. at the critical number x = the function f is concave up and corresponds to a relative minimum of f(x)
d. at the critical number x = the function f is concave down and corresponds to a relative minimum of f(x)
e. no conclusion can be made using the second - derivative test at this critical number

Explanation:

Step1: Find the first - derivative

Given $f(x)=-3x^{3}+3x^{2}-x - 5$, using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $f'(x)=-9x^{2}+6x - 1$.

Step2: Find the critical numbers

Set $f'(x) = 0$. So, $-9x^{2}+6x - 1=0$. This is a quadratic equation $ax^{2}+bx + c = 0$ with $a=-9$, $b = 6$, $c=-1$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, we get $x=\frac{-6\pm\sqrt{6^{2}-4\times(-9)\times(-1)}}{2\times(-9)}=\frac{-6\pm\sqrt{36 - 36}}{-18}=\frac{1}{3}$.

Step3: Find the second - derivative

Differentiate $f'(x)=-9x^{2}+6x - 1$ with respect to $x$. Using the power rule, $f''(x)=-18x + 6$.

Step4: Evaluate the second - derivative at the critical number

Substitute $x = \frac{1}{3}$ into $f''(x)$. So, $f''(\frac{1}{3})=-18\times\frac{1}{3}+6=-6 + 6=0$.

Step5: Apply the second - derivative test

When $f''(c)=0$ at a critical number $c$, the second - derivative test is inconclusive.

Answer:

E. No conclusion can be made using the second derivative test at this critical number