Question

the graph of the function has one relative maximum and one relative minimum point. plot these two points and check the concavity there. using only this information, sketch the graph.
f(x)=\frac{1}{3}x^{3}+3x^{2}+5x
the relative minimum point on the graph is .
(type an ordered pair, using integers or fractions.)
since the value of f at this relative minimum point is , at this point, the graph is
the relative maximum point on the graph is .
(type an ordered pair, using integers or fractions.)
since the value of f at this relative maximum point is , at this point, the graph is
which of the following graphs is f(x)?
a. image of graph a
b. image of graph b
c. image of graph c
d. image of graph d

Explanation:

Step1: Find the first - derivative

Differentiate $f(x)=\frac{1}{3}x^{3}+3x^{2}+5x$ using the power rule. $f'(x)=x^{2}+6x + 5$.

Step2: Find the critical points

Set $f'(x)=0$. So $x^{2}+6x + 5=(x + 1)(x + 5)=0$. The critical points are $x=-1$ and $x=-5$.

Step3: Find the second - derivative

Differentiate $f'(x)$ to get $f''(x)=2x+6$.

Step4: Determine relative extrema

For $x=-5$, $f''(-5)=2\times(-5)+6=-4<0$, so $x = - 5$ is a relative maximum. $f(-5)=\frac{1}{3}(-5)^{3}+3(-5)^{2}+5(-5)=\frac{-125}{3}+75 - 25=\frac{-125 + 225 - 75}{3}=\frac{25}{3}$. The relative maximum point is $(-5,\frac{25}{3})$.
For $x=-1$, $f''(-1)=2\times(-1)+6 = 4>0$, so $x=-1$ is a relative minimum. $f(-1)=\frac{1}{3}(-1)^{3}+3(-1)^{2}+5(-1)=\frac{-1 + 9-15}{3}=-\frac{7}{3}$. The relative minimum point is $(-1,-\frac{7}{3})$.

Step5: Analyze concavity

At the relative minimum point $x=-1$, $f''(-1)=4>0$, the graph is concave up.
At the relative maximum point $x=-5$, $f''(-5)=-4<0$, the graph is concave down.

Answer:

The relative minimum point on the graph is $(-1,-\frac{7}{3})$.
Since the value of $f''$ at this relative minimum point is $4$, at this point, the graph is concave up.
The relative maximum point on the graph is $(-5,\frac{25}{3})$.
Since the value of $f''$ at this relative maximum point is $-4$, at this point, the graph is concave down.
(To determine which graph is $f(x)$ among A, B, C, D, one would need to visually check which graph has a relative minimum at $(-1,-\frac{7}{3})$ (concave up there) and a relative maximum at $(-5,\frac{25}{3})$ (concave down there). Without seeing the exact scales and details of the graphs, we have completed the non - graphical parts of the problem.)