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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
[tex]-x^3+5x^2-2x-8=0[/tex][tex]\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}[/tex]Next, we find the result of :
[tex]\frac{f(x)}{x+1}[/tex][tex]\begin{gathered} So \\ \frac{-x^3+5x^2-2x-8}{x+1}=-x^2+6x-8 \end{gathered}[/tex]Now we solve
[tex]\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}[/tex]So the zeros of f(x) are -1, 2, and 4
Next, we find the stationary points.
[tex]\begin{gathered} \frac{df(x)}{dx}=-3x^2+10x-2 \\ \text{When }\frac{df(x)}{dx}=0,\text{ we have} \end{gathered}[/tex][tex]\begin{gathered} -3x^2+10x-2=0 \\ \text{Dividing through by -3 we have} \\ x^2-\frac{10}{3}x+\frac{2}{3}=0 \end{gathered}[/tex][tex]\begin{gathered} (x-\frac{5}{3})^2-(-\frac{5}{3})^2-2=0 \\ \Rightarrow(x-\frac{5}{3})^2=2+\frac{25}{9}=\frac{43}{9} \\ \Rightarrow x=\frac{5\pm\sqrt[]{43}}{3} \\ \Rightarrow x=2.55\text{ or }0.78 \end{gathered}[/tex][tex]\frac{d\frac{df(x)}{dx}}{dx}=-6x+10[/tex]At x = 2.55
[tex]\frac{d\frac{df(x)}{dx}}{dx}=-6(2.55)+10=-5.3<0[/tex]Hence we have a maximum point at x = 2.55
[tex]\frac{d\frac{df(x)}{dx}}{dx}=-6(0.78)+10=5.32>0[/tex]Hence, there is a minimum point at x = 0.78
[tex]\begin{gathered} f(0.78)\text{ = }-6.99 \\ f(2.550=2.83 \end{gathered}[/tex][tex]\begin{gathered} To\text{ check the intervals where the function increasing or decreasing} \\ \text{For x < 0.78 } \\ \frac{df(x)}{dx}=(x-0.78)(x-2.83)\text{ is positive} \\ \text{For 0.78 < x < 2.83 } \\ \frac{df(x)}{dx}=(x-0.78)(x-2.83)\text{ is negative} \\ \text{For x > 2.83} \\ \frac{df(x)}{dx}=(x-0.78)(x-2.83)\text{ is positive} \end{gathered}[/tex]This implies that
f is increasing on the intervals (–∞, 1/3) and (3, ∞)
