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.
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.
$\displaystyle f(x)=\sqrt{x}$
$\displaystyle g(x)=2^x$
$\displaystyle f^{-1}(x)\rightarrow x=\sqrt{y} \rightarrow y=x^2$
$\displaystyle \left[f^{-1} \circ g\right](x)=(2^x)^2=2^{2x}$
$\displaystyle 2^{2x}=.25$
$\displaystyle 2^{2x}=\frac{1}{4}$
$\displaystyle 2^{2x}=\frac{1}{2^2}$
$\displaystyle 2^{2x}=2^{-2}$
$\displaystyle 2x=-2$
$\displaystyle x=-1$
$\displaystyle f(x) = \sqrt{x}$
$\displaystyle y = \sqrt{x}$
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???