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 (( - 1,-\frac{7}{3})) (type an ordered pair, using integers or fractions.)
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 (square) (type an ordered pair, using integers or fractions.)

Explanation:

Step1: Find the first - derivative

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

Step2: Set the first - derivative equal to zero

$x^{2}+6x + 5 = 0$. Factoring the quadratic equation, we get $(x + 1)(x+5)=0$. So $x=-1$ or $x=-5$.

Step3: Find the second - derivative

Differentiate $f'(x)=x^{2}+6x + 5$ with respect to $x$. Using the power - rule, $f''(x)=2x + 6$.

Step4: Determine the nature of the critical points

When $x=-1$, $f''(-1)=2(-1)+6 = 4>0$, so $x = - 1$ is a relative minimum. When $x=-5$, $f''(-5)=2(-5)+6=-4<0$, so $x=-5$ is a relative maximum.

Step5: Find the $y$ - value of the relative maximum

Substitute $x=-5$ into $f(x)=\frac{1}{3}x^{3}+3x^{2}+5x$.
$f(-5)=\frac{1}{3}(-5)^{3}+3(-5)^{2}+5(-5)=\frac{-125}{3}+75 - 25=\frac{-125}{3}+50=\frac{-125 + 150}{3}=\frac{25}{3}$.

Answer:

$(-5,\frac{25}{3})$