Question

2-88. use a graphing calculator to graph $f(x)=\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-2x$. describe where the function is increasing, decreasing, concave up, concave down, and include any maxima or minima. homework help 2-89. algebraically solve each system of equations and state the method you used. check your answer graphically. homework help a. $4y - 5x=-2$ $x - 2y=-8$ b. $y =-\frac{3}{4}x + 9$ $2x - y = 2$

Explanation:

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Step1: Find the first - derivative

The function is $f(x)=\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-2x$. Using the power rule $(x^n)' = nx^{n - 1}$, we have $f'(x)=x^{2}-x - 2$.

Step2: Find critical points

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

Step3: Determine intervals of increase and decrease

Test intervals $(-\infty,-1)$, $(-1,2)$ and $(2,\infty)$.
For $x<-1$, let $x=-2$, then $f'(-2)=(-2)^{2}-(-2)-2=4 + 2-2=4>0$, so $f(x)$ is increasing on $(-\infty,-1)$.
For $-1For $x>2$, let $x = 3$, then $f'(3)=3^{2}-3 - 2=9-3 - 2=4>0$, so $f(x)$ is increasing on $(2,\infty)$.

Step4: Find the second - derivative

$f''(x)=2x - 1$.

Step5: Find inflection point

Set $f''(x)=0$, so $2x - 1=0$, then $x=\frac{1}{2}$.

Step6: Determine concavity

Test intervals $(-\infty,\frac{1}{2})$ and $(\frac{1}{2},\infty)$.
For $x<\frac{1}{2}$, let $x = 0$, then $f''(0)=-1<0$, so $f(x)$ is concave down on $(-\infty,\frac{1}{2})$.
For $x>\frac{1}{2}$, let $x = 1$, then $f''(1)=2\times1 - 1=1>0$, so $f(x)$ is concave up on $(\frac{1}{2},\infty)$.

Step7: Find maxima and minima

Since $f(x)$ changes from increasing to decreasing at $x=-1$, $f(-1)=\frac{1}{3}(-1)^{3}-\frac{1}{2}(-1)^{2}-2(-1)=-\frac{1}{3}-\frac{1}{2}+2=\frac{-2 - 3 + 12}{6}=\frac{7}{6}$ is a local maximum.
Since $f(x)$ changes from decreasing to increasing at $x = 2$, $f(2)=\frac{1}{3}(2)^{3}-\frac{1}{2}(2)^{2}-2(2)=\frac{8}{3}-2 - 4=\frac{8 - 6 - 12}{3}=-\frac{10}{3}$ is a local minimum.

Step1: Solve the system using the elimination method

We have the system

$$\begin{cases}4y-5x=-2\\x - 2y=-8\end{cases}$$

. Multiply the second equation by 2: $2x-4y=-16$.
Add this new equation to the first equation: $(4y-5x)+(2x - 4y)=-2+( - 16)$.
Simplifying gives $-3x=-18$, so $x = 6$.

Step2: Find the value of $y$

Substitute $x = 6$ into the second equation $6-2y=-8$.
Subtract 6 from both sides: $-2y=-8 - 6=-14$, then $y = 7$.

Step3: Check graphically

The first equation $4y-5x=-2$ can be rewritten as $y=\frac{5x - 2}{4}$. The second equation $x - 2y=-8$ can be rewritten as $y=\frac{x + 8}{2}$. Graphing these two lines, they should intersect at the point $(6,7)$.

Step1: Solve the system using the substitution method

We have the system

$$\begin{cases}y=-\frac{3}{4}x + 9\\2x-y=2\end{cases}$$

.
Substitute $y=-\frac{3}{4}x + 9$ into the second equation: $2x-(-\frac{3}{4}x + 9)=2$.
Expand to get $2x+\frac{3}{4}x-9 = 2$.
Combine like - terms: $\frac{8x+3x}{4}=2 + 9$, $\frac{11x}{4}=11$.
Multiply both sides by $\frac{4}{11}$, then $x = 4$.

Step2: Find the value of $y$

Substitute $x = 4$ into $y=-\frac{3}{4}x + 9$, then $y=-\frac{3}{4}\times4+9=-3 + 9=6$.

Step3: Check graphically

The first equation is $y=-\frac{3}{4}x + 9$ and the second equation $y=2x - 2$. Graphing these two lines, they should intersect at the point $(4,6)$.

Answer:

Increasing: $(-\infty,-1)\cup(2,\infty)$; Decreasing: $(-1,2)$; Concave up: $(\frac{1}{2},\infty)$; Concave down: $(-\infty,\frac{1}{2})$; Local maximum at $(-1,\frac{7}{6})$; Local minimum at $(2,-\frac{10}{3})$

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