Personally, on neither of those would I take the derivative. Instead, complete the square!
$2x^2- 8x+ 6= 2(x^2- 4x)+ 6= 2(x^2- 4x+ 4- 4)+ 6= 2(x- 2)^2- 8$.
Since a square is never the minimum is -8 when x= 2. We also need to check the endpoint x= 0. $2(0^2)- 8(0)= 6= 6$ and as x goes to infinity, the value increases indefinitely.
The absolute minimum is -8 and there is no absolute maximum.
$x^2$ is already a perfect square. Its minimum is 0 at x= 0. Then we need to check the endpoints of the intervals.
A. [-1, 1] $(-1)^2= 1$, and $1^2= 1$, The interval includes x= 0 so the minimum is 0 and the maximum is 1.
B. [1, 5] $1^2= 1$ and $5^2= 25$. The minimum is 1 and the maximum is 25.
C. [-5, 5] $(-5)^2= 26$ and $5^2= 25$. The interval includes 0 so the minimum is 0 and the maximum is 25.