3Part 1.
The critical values occur when
[tex]g^{\prime}(x)=0[/tex]In our case, we have
[tex]\frac{x^2-16}{x-2}=0[/tex]Then, for x different from 2, this last equation yields
[tex]\begin{gathered} x^2-16=0 \\ or\text{ equivalently} \\ (x+4)(x-4)=0 \end{gathered}[/tex]Therefore, the answer for part 1 is: x= 4 and x=-4
Part 2.
In order to see if the last points correspond to a minimum or maximim, we need to compute the seconde derivative of g. It is given by
[tex]\begin{gathered} g^{\doubleprime}(x)=\frac{(x-2)(2x)-(x^2-16)}{(x-2)^2} \\ or \\ g^{\doubleprime}(x)=\frac{(x-2)(2x)-(x+4)(x-4)}{(x-2)^2} \end{gathered}[/tex]Now, lets substitute the value x=4. It yields
[tex]\begin{gathered} g^{\doubleprime}(4)=\frac{(4-2)(2\cdot4)-(4+4)(4-4)}{(4-2)^2} \\ g^{\doubleprime}(4)=\frac{(2)(8)-0}{4} \\ g^{\doubleprime}(4)=4 \end{gathered}[/tex]and by substituting x= -4, we have
[tex]\begin{gathered} g^{\doubleprime}(-4)=\frac{(-4-2)(2\cdot(-4))-(-4+4)(-4-4)}{(-4-2)^2} \\ g^{\doubleprime}(-4)=\frac{(-6)(-8)-0}{(-6)^2} \\ g^{\doubleprime}(-4)=\frac{48}{36}=\frac{24}{18}=\frac{12}{9}=\frac{4}{3} \end{gathered}[/tex]Since both solutions are greater than zero, they both correspond to a local minimum points.
Part 3.
From the last result, we can see that, on the interval
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