Question

Sketch a graph of the polynomial function f(x) = –x3 + 5x2 – 2x – 8. Use the graph to complete the following:f is __________ on the intervals (–∞, 1/3) and (3, ∞)f is __________ on the interval (1/3, 3)f is __________ on the intervals (–∞, -1) and (2, 4)

Answer 1

To sketch the graph, we need to find the x-intercepts and y-intercepts.

To find the x-intercepts we solve the equation when y = 0.

That is

`-x^3+5x^2-2x-8=0`
`\begin{gathered} f(-1)=-(-1)^3+5(-1)^2-2(-1)-8=1+5+2-8=0 \\ \text{Hence} \\ x=-1\text{ is a zero of f(x)} \\ \Rightarrow\text{ x+1 is a factor of f(x)} \end{gathered}`

Next, we find the result of :

`(f(x))/(x+1)`
`\begin{gathered} So \\ (-x^3+5x^2-2x-8)/(x+1)=-x^2+6x-8 \end{gathered}`

Now we solve

`\begin{gathered} -x^2+6x-8=0 \\ \Rightarrow-x^2+2x+4x-8=0 \\ \Rightarrow-x(x-2)+4(x-2)=0 \\ \Rightarrow(4-x)(x-2)=0 \\ \Rightarrow x=4\text{ or 2} \end{gathered}`

So the zeros of f(x) are -1, 2, and 4

Next, we find the stationary points.

`\begin{gathered} (df(x))/(dx)=-3x^2+10x-2 \\ \text{When }(df(x))/(dx)=0,\text{ we have} \end{gathered}`
`\begin{gathered} -3x^2+10x-2=0 \\ \text{Dividing through by -3 we have} \\ x^2-(10)/(3)x+(2)/(3)=0 \end{gathered}`
`\begin{gathered} (x-(5)/(3))^2-(-(5)/(3))^2-2=0 \\ \Rightarrow(x-(5)/(3))^2=2+(25)/(9)=(43)/(9) \\ \Rightarrow x=\frac{5\pm\sqrt[]{43}}{3} \\ \Rightarrow x=2.55\text{ or }0.78 \end{gathered}`
`(d(df(x))/(dx))/(dx)=-6x+10`

At x = 2.55

`(d(df(x))/(dx))/(dx)=-6(2.55)+10=-5.3<0`

Hence we have a maximum point at x = 2.55

`(d(df(x))/(dx))/(dx)=-6(0.78)+10=5.32>0`

Hence, there is a minimum point at x = 0.78

`\begin{gathered} f(0.78)\text{ = }-6.99 \\ f(2.550=2.83 \end{gathered}`
`\begin{gathered} To\text{ check the intervals where the function increasing or decreasing} \\ \text{For x < 0.78 } \\ (df(x))/(dx)=(x-0.78)(x-2.83)\text{ is positive} \\ \text{For 0.78 < x < 2.83 } \\ (df(x))/(dx)=(x-0.78)(x-2.83)\text{ is negative} \\ \text{For x > 2.83} \\ (df(x))/(dx)=(x-0.78)(x-2.83)\text{ is positive} \end{gathered}`

This implies that

f is increasing on the intervals (–∞, 1/3) and (3, ∞)