Hello,

I can't solve this maths question and I would really appreciate your help.

My problem is:

Let f(x) =

__ (this means "square root of x" - I didn't know how to type it!)

V x

and g(x) = 2^x. Solve the equation

(f^-1 o g)(x) = 0.25.

Thanks.

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- Nov 29th 2008, 09:31 AMAminekhadirFunction notations/inverse functions.
Hello,

I can't solve this maths question and I would really appreciate your help.

My problem is:

Let f(x) =

__ (this means "square root of x" - I didn't know how to type it!)

V x

and g(x) = 2^x. Solve the equation

(f^-1 o g)(x) = 0.25.

Thanks. - Nov 29th 2008, 09:56 AMmasters
$\displaystyle f(x)=\sqrt{x}$

$\displaystyle g(x)=2^x$

$\displaystyle f^{-1}(x)\rightarrow x=\sqrt{y} \rightarrow y=x^2$

$\displaystyle \left[f^{-1} \circ g\right](x)=(2^x)^2=2^{2x}$

$\displaystyle 2^{2x}=.25$

$\displaystyle 2^{2x}=\frac{1}{4}$

$\displaystyle 2^{2x}=\frac{1}{2^2}$

$\displaystyle 2^{2x}=2^{-2}$

$\displaystyle 2x=-2$

$\displaystyle x=-1$ - Nov 29th 2008, 10:35 AMShyam
$\displaystyle f(x) = \sqrt{x}$

$\displaystyle y = \sqrt{x}$

For $\displaystyle f^{-1}(x)$, interchange, x and y and solve for y

$\displaystyle x = \sqrt{y}$

$\displaystyle x^2=y$

$\displaystyle y= x^2$

$\displaystyle f^{-1}(x)=x^2\;\;\;\;and\;\;\;\;g(x)=2^x$

Now,

$\displaystyle (f^{-1}\;\;o\;\;g)(x)=(2^x)^2$

$\displaystyle (2^x)^2=0.25$

$\displaystyle 2^{2x}=2^{-2}$

$\displaystyle (because,\;\; 0.25=\frac{1}{4}=\frac{1}{2^2}=2^{-2})$

$\displaystyle 2x=-2$

$\displaystyle x=-1$

did you get it now??? - Nov 29th 2008, 10:37 AMAminekhadir
Yes, thank you very much :)