Explanation
Question 57
[tex]\:f\left(x\right)=2x^3-15x^2+24x[/tex]To find the extreme values
[tex]\begin{gathered} \mathrm{Suppose\:that\:}x=c\mathrm{\:is\:a\:critical\:point\:of\:}f\left(x\right)\mathrm{\:then,\:} \\ \mathrm{If\:}f\:'\left(x\right)>0\mathrm{\:to\:the\:left\:of\:}x=c\mathrm{\:and\:}f\:'\left(x\right)<0\mathrm{\:to\:the\:right\:of\:}x=c\mathrm{\:then\:}x=c\mathrm{\:is\:a\:local\:maximum.} \\ \mathrm{If\:}f\:'\left(x\right)<0\mathrm{\:to\:the\:left\:of\:}x=c\mathrm{\:and\:}f\:'\left(x\right)>\:0\mathrm{\:to\:the\:right\:of\:}x=c\mathrm{\:then\:}x=c\mathrm{\:is\:a\:local\:minimum.} \\ \mathrm{If\:}f\:'\left(x\right)\mathrm{\:is\:the\:same\:sign\:on\:both\:sides\:of\:}x=c\mathrm{\:then\:}x=c\mathrm{\:is\:neither\:a\:local\:maximum\:nor\:a\:local\:minimum.} \end{gathered}[/tex]So, we will have the steps below
Step 1:
[tex]\begin{gathered} \mathrm{Plug\:the\:extreme\:point}\:x=0\:\mathrm{into}\:2x^3-15x^2+24x\quad \Rightarrow \quad \:y=0 \\ \mathrm{Minimum}\left(0,\:0\right) \end{gathered}[/tex]Step2:
[tex]\begin{gathered} \mathrm{Plug\:the\:extreme\:point}\:x=1\:\mathrm{into}\:2x^3-15x^2+24x\quad \Rightarrow \quad \:y=11 \\ \mathrm{Maximum}\left(1,\:11\right) \end{gathered}[/tex]Step 3:
[tex]\begin{gathered} \mathrm{Plug\:the\:extreme\:point}\:x=4\:\mathrm{into}\:2x^3-15x^2+24x\quad \Rightarrow \quad \:y=-16 \\ \mathrm{Minimum}\left(4,\:-16\right) \end{gathered}[/tex]Step 4:
[tex]\begin{gathered} \mathrm{Plug\:the\:extreme\:point}\:x=5\:\mathrm{into}\:2x^3-15x^2+24x\quad \Rightarrow \quad \:y=-5 \\ \mathrm{Maximum}\left(5,\:-5\right) \\ \end{gathered}[/tex]Thus, we will have
[tex]\mathrm{Minimum}\left(0,\:0\right),\:\mathrm{Maximum}\left(1,\:11\right),\:\mathrm{Minimum}\left(4,\:-16\right),\:\mathrm{Maximum}\left(5,\:-5\right)[/tex]Hence, our answer is
[tex]\begin{gathered} \begin{equation*} \mathrm{Minimum}\left(4,\:-16\right) \end{equation*} \\ \begin{equation*} \mathrm{Maximum}\left(1,\:11\right) \end{equation*} \end{gathered}[/tex]