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What is the value of the range of the function f(x)=x^2+2 for the domain value of 1/4?
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A. y=2 B. y = 2 1/4 C. y = 2 1/8 D. y = 2 1/16
domain represents the value that \(f(x)\) can take and range represents the value that \(x\) can take and we are given that domain is \(\large \frac{1}{4}\) so therefore -> \(\large f(x)=x^2+2\) we have \(f(x)=\large \frac{1}{4}\) so put it in the equation we get this- \(\large \frac{1}{4}=x^2+2\) now try to find the value of \(x\) from here :) that will be your range
Wait, I believe it is y = 2 1/16 or D
Is D correct?
Yes.
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Ok, thank you :)
Thanks for asking :-)
@skullpatrol but solving that isn't giving real values of x :|
You have to solve for y given x=1/4.
oh kay i said the opposite damn
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