How to find $f(x)$ from $f(g(x))$, $g(x)$, ex:
$g(x)=1-x^2$
$f(g(x))=(x^4+x^2)/(1+x^2)$
$f(1/2)=$
use matlab code....I don't know..help me thanks a lot!
How to find $f(x)$ from $f(g(x))$, $g(x)$, ex:
$g(x)=1-x^2$
$f(g(x))=(x^4+x^2)/(1+x^2)$
$f(1/2)=$
use matlab code....I don't know..help me thanks a lot!
I give you an example, and hope you can translate it to Matlab. I don't know Matlab much but I think you can do it by yourself.
Assume you have $h(x)=f(g(x))=4x^2+4x+7$ and $g(x)=2x+1$ and want to find $f(x)$. Define $h(x)$ and $g(x)$ in Matlab environment and solve $g(x)=t$ respect to $x$(which $t$ is a parameter) You get: $x=0.5t-0.5$ Now put what you got into $h(x)$, so you get: $h(0.5t-0.5)=t^2+6$ This is your desired function $f(x)$ if you replace $t$ to $x$.
For $g(x) = 1 - x^2$ to be equal to $1/2$, $x = 1/\sqrt{2}$.
So: $f(1/2) = \frac{1/4 + 1/2}{1 + 1/2} = 0.75/1.5 = 1/2$
A simple solution is to notice that $f(g(x)) = \dfrac{x^4+x^2}{x^2+1} = \dfrac{x^2(x^2+1)}{x^2+1}= x^2 = 1 - g(x)$ so if $g(x)=\frac{1}{2}$ then $f(g(x))=1-\frac{1}{2} = \frac{1}{2}$.
clc; syms f g x; S = solve('g = 1 - x ^ 2', 'g = 1 / 2') x = S.x(1); f = (x ^ 4 + x ^ 2) / (1 + x ^ 2); double (f)