y = 2^[(x^2) - 2x]
I need solution!!
Thanks in advance
The question is: What is the minimum value of the function f(x) = 2^[(x^2) - 2x] ?
We have to complete the square and then find the minimum value for y.
im stuck here:
y = 2 ^ [ ((x-1)^2) -1 ] , now how can i get rid of the 2?
the answer is 2^(-1), i want the solution
sry about all the brackets
You should know what an exponent is right? if not we have a problem.
So the question I asked you is what when we find the minimum value of
$\displaystyle x^2-2x$ why does this make
$\displaystyle 2^{x^2-2x}$ a minimum?
Completing the square gives
$\displaystyle x^2-2x=x^2-2x+1-1=(x-1)^2-1$
The minimum value occurs when the term $\displaystyle (x-1)^2=0$
This gives the smallest value of the quadratic as $\displaystyle -1$
So the minimum function value is $\displaystyle 2^{-1}=\frac{1}{2}$