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}+2x^{2}+3x
the relative minimum point on the graph is (-1, -\frac{4}{3})
(type an ordered pair, using integers or fractions.)
since the value of f at this relative minimum point is 2, at this point, the graph is concave up.
the relative maximum point on the graph is (-3,0)
(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

Explanation:

Step1: Find the first - derivative of \(f(x)\)

Given \(f(x)=\frac{1}{3}x^{3}+2x^{2}+3x\), then \(f'(x)=x^{2}+4x + 3\).

Step2: Find the critical points

Set \(f'(x)=0\), so \(x^{2}+4x + 3=(x + 1)(x+3)=0\). The critical points are \(x=-1\) and \(x=-3\).

Step3: Find the second - derivative of \(f(x)\)

Differentiate \(f'(x)=x^{2}+4x + 3\) with respect to \(x\), we get \(f''(x)=2x + 4\).

Step4: Evaluate \(f''(x)\) at the critical points

When \(x=-3\), \(f''(-3)=2\times(-3)+4=-2\).

Answer:

Since the value of \(f''\) at the relative maximum point \(x = - 3\) is \(-2\), at this point, the graph is concave down.