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 blank at this point, the graph is blank

Explanation:

Step1: Find the first - derivative

First, find the derivative of \(f(x)=\frac{1}{3}x^{3}+2x^{2}+3x\). Using the power rule \((x^n)^\prime = nx^{n - 1}\), we have \(f^\prime(x)=x^{2}+4x + 3\).

Step2: Find the critical points

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

Step3: Find the second - derivative

Differentiate \(f^\prime(x)\) to get \(f^{\prime\prime}(x)=2x + 4\).

Step4: Evaluate the second - derivative at the relative minimum point

We know the relative minimum point is \(x=-1\). Substitute \(x = - 1\) into \(f^{\prime\prime}(x)\): \(f^{\prime\prime}(-1)=2\times(-1)+4=2\). Since \(f^{\prime\prime}(-1)>0\), the graph is concave up at the point \((-1,-\frac{4}{3})\).

Answer:

2, concave up