Question

57. do not use the answer under the line in the explanation itself, only refer to it to make sure of your work. USE DERIVITIVES NOT GRAPHING

Answer 1

Step-by-step explanation

Question 57

`\:f\left(x\right)=2x^3-15x^2+24x`

To find the extreme values

So, we will have the steps below

Step 1:

`\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}`

Step2:

`\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}`

Step 3:

`\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}`

Step 4:

`\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}`

Thus, we will have

`\mathrm{Minimum}\left(0,\:0\right),\:\mathrm{Maximum}\left(1,\:11\right),\:\mathrm{Minimum}\left(4,\:-16\right),\:\mathrm{Maximum}\left(5,\:-5\right)`

Hence, our answer is

`\begin{gathered} \begin{equation*} \mathrm{Minimum}\left(4,\:-16\right) \end{equation*} \\ \begin{equation*} \mathrm{Maximum}\left(1,\:11\right) \end{equation*} \end{gathered}`