Question

URGENTEEEE
DOU 30 PONTOS

Answer 1

a) f(x) = -4x + 1

y = -4x + 1

Função inversa:

x = -4y + 1
4y = -x + 1
$y \ = \ \frac{-x \ + \ 1}{4}$

$\boxed{\bold{f^{-1}(x) \ = \ \frac{-x \ + \ 1}{4}}}$



b) $f(x) \ = \ \sqrt[5]{x \ + \ 3}$

$y \ = \ \sqrt[5]{x \ + \ 3}$

Função inversa:

$x \ = \ \sqrt[5]{y \ + \ 3}$
$x \ = \ (y \ + \ 3)^{\frac{1}{5}}$
$(y \ + \ 3)^{\frac{1}{5}} \ = \ x$
$y \ + \ 3 \ = \ x^5$
$y \ = \ x^5 \ - \ 3$

$\boxed{\bold{f^{-1}(x) \ = \ x^5 \ - \ 3}}$



c) $f(x) \ = \ \frac{x}{2} \ - \ 5$

$y \ = \ \frac{x}{2} \ - \ 5$

Função inversa:

$x \ = \ \frac{y}{2} \ - \ 5$
$x \ + \ 5 \ = \ \frac{y}{2}$
$\frac{y}{2} \ = \ x \ + \ 5$
$y \ = \ 2 . \ (x \ + \ 5)$
$y \ = \ 2x \ + \ 10$

$\boxed{\bold{f^{-1}(x) \ = \ 2x \ + \ 10}}$



d) $f(x) \ = \ \frac{2}{x \ - \ 1}$

$y \ = \ \frac{2}{x \ - \ 1}$

Função inversa:

$x \ = \ \frac{2}{y \ - \ 1}$
$x \ . \ (y \ - \ 1) = \ 2$
$(y \ - \ 1) = \ \frac{2}{x}$
$y \ - \ 1 = \ \frac{2}{x}$
$y \ = \ \frac{2}{x} \ + \ 1$

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



e) $f(x) \ = \ 3x^2 \ + \ 4$

$y \ = \ 3x^2 \ + \ 4$

Função inversa:

$x \ = \ 3y^2 \ + \ 4$
$x \ - \ 4 \ = \ 3y^2$
$3y^2 \ = \ x \ - \ 4$
$y^2 \ = \ \frac{x \ - \ 4}{3}$
$y \ = \ \sqrt{\frac{x \ - \ 4}{3}}$

$\boxed{\bold{f^{-1}(x) \ = \ \sqrt{\frac{x \ - \ 4}{3}}}}$



f) $f(x) \ = \ \frac{6x \ - \ 1}{3x \ + \ 2}$

$y \ = \ \frac{6x \ - \ 1}{3x \ + \ 2}$ 

Função inversa:

$x \ = \ \frac{6y \ - \ 1}{3y \ + \ 2}$ 

$x \ . \  (3y \ + \ 2) \ = \ 6y \ - \ 1$ 
$3xy \ + \ 2x \ = \ 6y \ - \ 1$ 
$3xy \ + \ 2x \ - \ 6y \ + \ 1 \ = \ 0$
$3xy \ - \ 6y \ + \ 2x \ + \ 1 \ = \ 0$
$y \ . \ (3x \ - \ 6) \ = \ -2x \ - \ 1$

$y \ = \ \frac{-2x \ - \ 1}{3x \ - \ 6}$

$\boxed{\bold{f^{-1}(x) \ = \ \frac{-2x \ - \ 1}{3x \ - \ 6}}}$

Answer 2

As respostas:

a) $f(x) = -4x+1$

$x=-4y+1$

$x-1=-4y$

$y= \frac{x-1}{4}$

b) $y= \sqrt[5]{x+3}$

$x= \sqrt[5]{y+3}$

$x^{5} =y+3$

$y= x^{5}-3$

c) $y= \frac{x}{2} -5$

$x= \frac{y}{2} -5$

$\frac{y}{2} =x+5$

$y=2x+10$

d) $y= \frac{2}{x-1}$

$x= \frac{2}{y-1}$

$y-1= \frac{2}{x}$

$y=2x+1$

e) $y= 3x^{2} +4$

$x= 3y^{2} +4$

$x-4=3 y^{2}$

$y^{2} = \frac{x-4}{3}$

$y= \sqrt{ \frac{x-4}{3} }$

f) $y = \frac{6x-1}{3x+2}$

$x= \frac{6y-1}{3y+2}$

$3yx+2x=6y-1$

$3yx-6y+2x+1=0$

$y(3x-6)+2x+1=0$

$y= \frac{-2x-1}{3x-6}$