the graph of the function has one relative maximum and one relative minimum point. plot these two points and…

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 (1,-\frac{5}{3}) (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 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 (1,-\frac{5}{3}) (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

Answer

Explanation:

Step1: Find the first - derivative

Given (f(x)=\frac{1}{3}x^{3}+x^{2}-3x), using the power rule ((x^n)^\prime = nx^{n - 1}), we have (f^\prime(x)=x^{2}+2x - 3).

Step2: Find the critical points

Set (f^\prime(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^\prime(x)=x^{2}+2x - 3) with respect to (x). Using the power rule, (f^{\prime\prime}(x)=2x + 2).

Step4: Evaluate the second - derivative at the relative minimum point (x = 1)

Substitute (x = 1) into (f^{\prime\prime}(x)), (f^{\prime\prime}(1)=2\times1+2=4). Since (f^{\prime\prime}(1)=4>0), the graph is concave up at the point ((1,-\frac{5}{3})).

Answer:

The value of (f^{\prime\prime}) at this relative minimum point is (4), at this point, the graph is concave up.