Thread: Minimum value for the function:

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

I need solution!!

Thanks in advance

When does the exponential part of the equation have its minimum value?

is minimized when? Why will this solve your problem?

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

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.

Originally Posted by Sorena

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

why does this make

a minimum?

Completing the square gives



The minimum value occurs when the term

This gives the smallest value of the quadratic as

So the minimum function value is

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!))