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}+x^{2}-3x
the relative minimum point on the graph is ( left(1,-\frac{5}{3}
ight) )
(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 ( (-3,9) )
(type an ordered - pair, using integers or fractions.)
since the value of ( f ) at this relative maximum point is (square) at this point, the graph is (square)

Explanation:

Step1: Find the first - derivative

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

Step2: Find the critical points

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

Step3: Find the second - derivative

Differentiate $f'(x)$ with respect to $x$. $f''(x)=2x + 2$.

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

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

Answer:

The value of $f''$ at the relative maximum point is $-4$, at this point, the graph is concave down.