+635 Find the minimum value of 9^x - 2 \cdot 3^x + 1 over all real numbers x.
+129883 \(9^x - 2 \cdot 3^x + 1 \)
(3^x - 1)^2
Take the derivative and set to 0 {close your eyes if you haven't had Calculus!! }
2 (3^x -1) ( 3^x * log 3 ) = 0
(3^x -1) * 3^x = 0 { 3^x is always +] [divide this out ]
3^x -1 = 0
3^x =1
x = 0
The min is
9^0 - 2*3^0 + 1 = 1 - 2(1) + 1 = 0
