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Find the minimum value of 9^x - 2 \cdot 3^x + 1 over all real numbers x.

 Jun 10, 2024
 #1
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\(9^x - 2 \cdot 3^x + 1 \)

 

(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

 

 

 

 

 

cool cool cool

 Jun 10, 2024

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