Question

Dadas as funções de R em R, definidas por f(x)=2x, g(x)=3x+1 e h(x)=x2-1, calcule a)h(f(g(1))) b)f(g(h(1/2))) c)f(h(g(0))).​

Answer 1

Explicação passo-a-passo:

a) • $\sf g(1)=3\cdot1+1$

$\sf g(1)=3+1$

$\sf g(1)=4$

Assim, $\sf h(f(g(1)))=h(f(4))$

• $\sf f(4)=2\cdot4$

$\sf f(4)=8$

Então, $\sf h(f(g(1)))=h(8)$

• $\sf h(8)=8^2-1$

$\sf h(8)=64-1$

$\sf h(8)=63$

Logo, $\sf h(f(g(1)))=63$

b) • $\sf h\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}\right)^2-1$

$\sf h\left(\dfrac{1}{2}\right)=\dfrac{1}{4}-1$

$\sf h\left(\dfrac{1}{2}\right)=\dfrac{1-4}{4}$

$\sf h\left(\dfrac{1}{2}\right)=\dfrac{-3}{4}$

Assim, $\sf f\left(g\left(h\left(\dfrac{1}{2}\right)\right)\right)=f\left(g\left(\dfrac{-3}{4}\right)\right)$

• $\sf g\left(\dfrac{-3}{4}\right)=3\cdot\left(\dfrac{-3}{4}\right)+1$

$\sf g\left(\dfrac{-3}{4}\right)=\dfrac{-9}{4}+1$

$\sf g\left(\dfrac{-3}{4}\right)=\dfrac{-9+4}{4}$

$\sf g\left(\dfrac{-3}{4}\right)=\dfrac{-5}{4}$

Então, $\sf f\left(g\left(h\left(\dfrac{1}{2}\right)\right)\right)=f\left(\dfrac{-5}{4}\right)$

• $\sf f\left(\dfrac{-5}{4}\right)=2\cdot\left(\dfrac{-5}{4}\right)$

$\sf f\left(\dfrac{-5}{4}\right)=\dfrac{-10}{4}$

$\sf f\left(\dfrac{-5}{4}\right)=\dfrac{-5}{2}$

Logo, $\sf f\left(g\left(h\left(\dfrac{1}{2}\right)\right)\right)=\dfrac{-5}{2}$

c) • $\sf g(0)=3\cdot0+1$

$\sf g(0)=0+1$

$\sf g(0)=1$

Assim, $\sf f(h(g(0)))=f(h(1))$

• $\sf h(1)=1^2-1$

$\sf h(1)=1-1$

$\sf h(1)=0$

Então, $\sf f(h(g(0)))=f(0)$

• $\sf f(0)=2\cdot0$

$\sf f(0)=0$

Logo, $\sf f(h(g(0)))=0$