Question

Dada $f(x)=6x+2$;
$g(x)=4x-1$;
$h(x)=\frac {2x-3}{2}$;
calcular x de modo que: $f(h(x))+g^{-1}(f(x))=f(g(h(2)))+g(f^-1(8))$

Nota: Não estou conseguindo obter o resultado exato de x, a resposta é $\frac {23}{10}$
Matéria: Função Composta/Inversa

Answer 1

$\boxed{\boxed{f(x)=6x+2}}\\\\ x=6y+2\\\\\boxed{\boxed{f^{-1}(x)= \frac{x-2}{6} }}\\\\\\\boxed{\boxed{f^{-1}(8)= \frac{8-2}{6} =1}}$


.
$\boxed{\boxed{h(x)= \frac{2x-3}{2}}}\\\\\\\boxed{\boxed{h(2)= \frac{2*2-3}{2}= \frac{1}{2} }}$

.
$\boxed{\boxed{g(x)=4x-1}}\\\\\\\boxed{\boxed{g(f^{-1}(8))=g(1)=4*1-1=3 }}\\\\\\\boxed{\boxed{g(h(2))=g\left( \frac{1}{2} \right)= 4* \frac{1}{2}-1 =1}}$

.

$f(h(x)) = 6*( \frac{2x-3}{2})+2\\\\\\\boxed{\boxed{f(h(x))=6x-7}} \\\\\\\boxed{\boxed{f(g(h(2))=f(1)=6*1+2=8}}\\\\\\\\g^{-1}f(x)= \frac{(6x+2)+1}{4} \\\\\\\boxed{\boxed{g^{-1}f(x)= \frac{6x+3}{4} }}$


temos
$\bmatrix{f(h(x))=6x-7\\\\g^{-1}f(x)= \frac{6x+3}{4} \\\\f(g(h(2))=8\\\\g(f^{-1}(8))=3\end$

..
resolvendo
$6x-7 + \frac{6x+3}{4} = 8+3\\\\4*\left[6x-7 + \frac{6x+3}{4}\right] = 4*[11]\\\\24x-28+6x+3=44\\\\30x-25=44\\\\x= \frac{69}{30} = \frac{23}{10}$