In order to find a critical value (maximum or minimum) we need to compute the first derivative, which is given by
[tex]\frac{d}{dx}g(x)=6x-6[/tex]
Then, a critical value ocurrs when
[tex]\frac{d}{dx}g(x)=0[/tex]
which implies that
[tex]6x-6=0[/tex]
So by adding 6 to both side, we have
[tex]\begin{gathered} 6x=6 \\ then \\ x=\frac{6}{6} \\ x=1 \end{gathered}[/tex]
Therefore, there is a maximum or minimum at x=1.
In order to see if the point represents a maximum or minimum, we need to find the second derivative of our function, which is given by
[tex]\frac{d^2}{dx^2}g(x)=6[/tex]
We have that if the second derivative is positive, the point represents a minimum and if it is negative, the point represents a maximum. In our case, since the second derivative is greater than zero (positive number) there is a minimum point at x=1. Then, by substituting this values into the function, we get
[tex]\begin{gathered} g(1)=3(1^2)-6(1)+5 \\ g(1)=3-6+5 \\ g(1)=2 \end{gathered}[/tex]
so the minimum point is located at (1,2).
Therefore, with the above information, the answers are:
Does the function have a minimum or maximum values? Answer: Minimum
Where does the minimum or maximum value occur? Answer: x=1
Whats is the funtion's minimum or maximum values? Answer: 2
