Solution
We are given the function
[tex]f(x)=x^3-5x^2+8x[/tex]First, we will find the critical point. To find the critical points, we set f'(x) = 0
[tex]\begin{gathered} f(x)=x^{3}-5x^{2}+8x \\ By\text{ differentiating} \\ f^{\prime}(x)=3x^2-10x+8 \end{gathered}[/tex]Now, we set f'(x) = 0
[tex]\begin{gathered} f^{\prime}(x)=3x^{2}-10x+8 \\ 3x^2-10x+8=0 \\ (3x-4)(x-2)=0 \\ x=2,\frac{4}{3} \end{gathered}[/tex]So we have to split the interval into
[tex](-\infty,\frac{4}{3}),(\frac{4}{3},2),(2,\infty)[/tex]We check the intervals
[tex]\begin{gathered} On\text{ the interval }(-\infty,\frac{4}{3}) \\ f^{\prime}(x)\text{ is always positive} \\ Therefore,\text{ f is increasing on this interval} \end{gathered}[/tex]The next interval
[tex]\begin{gathered} On\text{ the interval }(\frac{4}{3},2) \\ f^{\prime}(x)\text{ is negative on this interval} \\ Therefore,\text{ f is decreasing on the interval} \end{gathered}[/tex]The last interval
[tex]\begin{gathered} On\text{ the interval }(2,\infty) \\ f^{\prime}(x)\text{ is positive on this interval} \\ Therefore,\text{ f is increasing on the interval} \end{gathered}[/tex]The graph of the function is given below
The function is increasing on the interval
[tex](-\infty,\frac{4}{3})\cup(2,\infty)[/tex]The function is decreasing on the interval
[tex](\frac{4}{3},2)[/tex]Local Minimum
[tex]\begin{gathered} f(2)=4 \\ Local\text{ minimum is 4} \\ x\text{ value is 2} \end{gathered}[/tex]Local Maximum
[tex]\begin{gathered} f(\frac{4}{3})=\frac{112}{27} \\ Local\text{ maximum is }\frac{112}{27} \\ x\text{ value is }\frac{4}{3} \end{gathered}[/tex]
