Find the minimum value of 9^x - 2 \cdot 3^x + 1 over all real numbers x.
Take the derivative and set to 0
ln 9 (9^x) - 2ln3* 3^x = 0
ln 9 ( 9^x) - ln 3^2 * 3^x = 0
ln9 (9^x) - ln 9 ( 3^x) = 0 divide out ln 9
9^x - 3^x = 0
9^x = 3^x
(9/3)^x = 1
3^x = 1
x = 0
y= 9^(0) - 2*3^(0) + 1 =
1 - 2 + 1 =
0 = is the min