Question

answer the questions for the function f(x)= - 3x^3 + 3x^2 - x - 5
at the critical number x = the function f is concave up and corresponds to a relative minimum of f(x).
at the critical number x = the function f is concave down and corresponds to a relative minimum of f(x).
no conclusion can be made using the second derivative test at this critical number
d. set the formula for the derivative f(x)=0 to find any possible inflection points
f(x)=0 at x = 1/3
use the table feature for the graph of f(x) to examine the concavity of the graph of f(x). what can you conclude? select the correct choice below and, if necessary, fill in the answer boxes within your choice
there is a point of inflection at x = where the graph of f changes from concave up to concave down. the slope of the graph of f at this inflection point is
there is a point of inflection at x = where the graph of f changes from concave down to concave up. the slope of the graph of f at this inflection point is
no conclusion can be made

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 second - derivative

Differentiate $f'(x)$ with respect to $x$. So $f''(x)=-18x + 6$.

Step3: Analyze the concavity and inflection points

Set $f''(x) = 0$, then $-18x+6 = 0$. Solving for $x$ gives $x=\frac{1}{3}$.
When $x<\frac{1}{3}$, let's take $x = 0$, then $f''(0)=6>0$, so the function is concave up on the interval $(-\infty,\frac{1}{3})$.
When $x>\frac{1}{3}$, let's take $x=\frac{2}{3}$, then $f''(\frac{2}{3})=-18\times\frac{2}{3}+6=-12 + 6=-6<0$, so the function is concave down on the interval $(\frac{1}{3},\infty)$.
The point of inflection occurs at $x = \frac{1}{3}$. To find the slope of the graph of $f$ at the inflection - point, we find $f'(\frac{1}{3})$.
$f'(\frac{1}{3})=-9\times(\frac{1}{3})^{2}+6\times\frac{1}{3}-1=-9\times\frac{1}{9}+2 - 1=-1 + 2 - 1=0$.

Answer:

B. There is a point of inflection at $x=\frac{1}{3}$ where the graph of $f$ changes from concave down to concave up. The slope of the graph of $f$ at this inflection point is $0$.