y = 2^[(x^2) - 2x]

I need solution!!

Thanks in advance

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- Nov 5th 2010, 06:19 PMSorenaMinimum value for the function:
y = 2^[(x^2) - 2x]

I need solution!!

Thanks in advance - Nov 5th 2010, 06:22 PMTheEmptySet
When does the exponential part of the equation have its minimum value?

$\displaystyle x^2-2x$ is minimized when? Why will this solve your problem? - Nov 5th 2010, 06:33 PMSorena
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 - Nov 5th 2010, 06:37 PMEducated
Like the previous question you asked in the geometry section, calculus is involved.

Differentiation is required for this one, when dy/dx = 0, x will be your minimum point. - Nov 5th 2010, 06:45 PMTheEmptySet
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 $\displaystyle 2^{-1}=\frac{1}{2}$ - Nov 5th 2010, 06:55 PMSorena
Thank you soooo much.

(sry i didnt mean to be rude earlier, I thought you're signature was your solution to my problem, lol (dumb I know!))