Question

identify the inflection points and local maxima and minima of the graphed function. identify the open intervals on which the function is differentiable and is concave up and concave down.

find each local maximum. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

a. there is one local maximum value of at

(simplify your answers.)

b. there are two local maxima. in increasing order of x - value, the values are and at and, respectively.

(simplify your answers.)

c. there are no local maxima

Explanation:

Step1: Find the first derivative

Differentiate \( y=\frac{x^{3}}{3}-x^{2}-3x \) using the power rule \( (x^n)^\prime = nx^{n - 1} \).
\( y^\prime=\frac{3x^{2}}{3}-2x - 3=x^{2}-2x - 3\).
Set \( y^\prime = 0 \), so \( x^{2}-2x - 3=(x - 3)(x+1)=0 \). The critical points are \( x=-1 \) and \( x = 3 \).

Step2: Find the second derivative

Differentiate \( y^\prime=x^{2}-2x - 3 \). \( y^{\prime\prime}=2x-2\).
Set \( y^{\prime\prime}=0 \), then \( 2x - 2=0\), which gives \( x = 1 \).

Step3: Use the second - derivative test for local maxima/minima

• For \( x=-1 \): \( y^{\prime\prime}(-1)=2\times(-1)-2=-4<0 \). So the function has a local maximum at \( x=-1 \).

Substitute \( x=-1 \) into \( y=\frac{x^{3}}{3}-x^{2}-3x \):
\( y=\frac{(-1)^{3}}{3}-(-1)^{2}-3\times(-1)=-\frac{1}{3}-1 + 3=\frac{-1 - 3+9}{3}=\frac{5}{3}\).

• For \( x = 3 \): \( y^{\prime\prime}(3)=2\times3-2=4>0 \). So the function has a local minimum at \( x = 3 \).

Answer:

A. There is one local maximum value of \(\frac{5}{3}\) at \(x=-1\).