Question

Dada as funções f(x)=5x-8, g(x)=x2+4x e h(x)=2/x-3 determine

a)f(g(x))
b)h(g(x))
c)f(h(x))
d)h(f(x))
e)g(f(x))
f)f(f(x))
g)g(g(x))
h) h(h(x))

Answer 1

¡Hey!!

Explicação passo-a-passo:

$f(x) = 5x - 8 \\ g(x) = {x}^{2} + 4x \\ h(x) = \frac{2}{x - 3}$

$a) \: f(g(x)) = 5.(g(x)) - 8 \\ = 5.( {x}^{2} + 4x) - 8 \\ = {5x}^{2} + 20x - 8$

$b) \: h(g(x)) = \frac{2}{x - 3}$

$b) \: h(g(x)) = \frac{2}{(g(x)) - 3} \\ = \frac{2}{ {x}^{2} + 4x - 3 }$

$c) \: f(h(x)) = 5.(h(x)) - 8 \\ = 5.( \frac{2}{x - 3} ) - 8 \\ = \frac{10}{x - 3} - 8$

$d) \: h(f(x)) = \frac{2}{(f(x)) - 3} \\ = \frac{2}{(5x - 8) - 3} \\ = \frac{2}{5x - 8 - 3} \\ = \frac{2}{5x - 11}$

$e) \: g(f(x)) = ({f(x)})^{2} + 4.(f(x)) \\ = ({5x - 8)}^{2} + 4.(5x - 8) \\ = (5x - 8).(5x - 8) + 20x - 32 \\ = {25x}^{2} - 45x - 45x + 64 + 20x - 32 \\ = {25x}^{2} - 90x + 64 + 20x - 32 \\ = {25x}^{2} - 90x + 20x + 64 - 32 \\ = {25x}^{2} - 70x + 42$

$f) \: f(f(x)) = 5.(f(x)) - 8 \\ = 5.(5x - 8) - 8 \\ = 25x - 45 - 8 \\ = 25x - 53$

$g(g(x)) = ( {g(x)})^{2} + 4.(g(x)) \\ = ( { {x}^{2} + 4x) }^{2} + 4.( {x}^{2} + 4x) \\ = ( {x}^{2} + 4x).( {x}^{2} + 4x) + 4 {x}^{2} + 16x \\ = {x}^{4} + 4 {x}^{3} + 4 {x}^{3} + 16 {x}^{2} + {4x}^{2} + 16x \\ = {x}^{4} + {8x}^{3} + {20x}^{2} + 16x$

$h) \: h(h(x) ) = \frac{2}{(h(x)) - 3} \\ = \frac{2}{( \frac{2}{x - 3} ) - 3} \\ = \frac{2}{ \frac{2}{x - 3} - 3 } \\ = \frac{2 \times (x - 3)}{ - 3 \times 2} \\ = \frac{2x - 6}{ - 6} \\ = - \frac{2x - 6}{6}$

¡Espero ter ajudado

- Bons estudos & Feliz Natal..