Question

answer the questions for the function f(x)= - 3x^3 + 3x^2 - x - 5
there is a point of inflection at x = 1/3 where the graph of f changes from concave up to concave down. the slope of the graph of f at this inflection point is 0
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.
choose the graph of the function f(x)

Explanation:

Step1: Find the first - derivative

Using the power rule $(x^n)'=nx^{n - 1}$, for $f(x)=-3x^{3}+3x^{2}-x - 5$, we have $f'(x)=-9x^{2}+6x - 1$.

Step2: Find the second - derivative

Differentiate $f'(x)$ again. $f''(x)=-18x + 6$.

Step3: Find the inflection point

Set $f''(x) = 0$, so $-18x+6 = 0$. Solving for $x$ gives $x=\frac{1}{3}$.

Step4: Find the slope at the inflection point

Substitute $x = \frac{1}{3}$ into $f'(x)$. $f'(\frac{1}{3})=-9(\frac{1}{3})^{2}+6(\frac{1}{3})-1=-9\times\frac{1}{9}+2 - 1=-1 + 2-1=0$.

Step5: Analyze concavity

For $x<\frac{1}{3}$, let's take $x = 0$, then $f''(0)=6>0$, so the function is concave up. For $x>\frac{1}{3}$, 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.

Answer:

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