Question

Seja f a função de R em R assim definida f(x) = x³ + 5x² + 2 se x ∈ Q x² + 2x + 1 se x não pertence a Q. Calcule f (-3/7) , f(√2), f(√4), f(√3 − 1) e f(0, 75).

Answer 1

Explicação passo-a-passo:

$\sf f(x)=\begin{cases} \sf x^3+5x^2+2,~se~x\in\mathbb{Q} \\ \sf x^2+2x+1=(x+1)^2,~se~x\not\in\mathbb{Q}\end{cases}$

a)

$\sf -\dfrac{3}{7}\in\mathbb{Q}$

$\sf f\left(-\dfrac{3}{7}\right)=\left(-\dfrac{3}{7}\right)^3+5\cdot\left(-\dfrac{3}{7}\right)^2+2$

$\sf f\left(-\dfrac{3}{7}\right)=-\dfrac{27}{343}+5\cdot\dfrac{9}{49}+2$

$\sf f\left(-\dfrac{3}{7}\right)=-\dfrac{27}{343}+\dfrac{45}{49}+2$

$\sf f\left(-\dfrac{3}{7}\right)=\dfrac{-27+315+686}{343}$

$\sf f\left(-\dfrac{3}{7}\right)=\dfrac{974}{343}$

b)

$\sf \sqrt{2}\not\in\mathbb{Q}$

$\sf f(\sqrt{2})=(\sqrt{2}+1)^2$

$\sf f(\sqrt{2})=(\sqrt{2})^2+2\cdot\sqrt{2}\cdot1+1^2$

$\sf f(\sqrt{2})=2+2\sqrt{2}+1$

$\sf f(\sqrt{2})=3+2\sqrt{2}$

c)

$\sf \sqrt{4}=2\in\mathbb{Q}$

$\sf f(\sqrt{4})=2^3+5\cdot2^2+2$

$\sf f(\sqrt{4})=8+5\cdot4+2$

$\sf f(\sqrt{4})=8+20+2$

$\sf f(\sqrt{4})=30$

d)

$\sf \sqrt{3}-1\not\in\mathbb{Q}$

$\sf f(\sqrt{3}-1)=(\sqrt{3}-1+1)^2$

$\sf f(\sqrt{3}-1)=(\sqrt{3})^2$

$\sf f(\sqrt{3}-1)=3$

e)

$\sf 0,75=\dfrac{3}{4}\in\mathbb{Q}$

$\sf f(0,75)=\left(\dfrac{3}{4}\right)^3+5\cdot\left(\dfrac{3}{4}\right)^2+2$

$\sf f(0,75)=\dfrac{27}{64}+5\cdot\dfrac{9}{16}+2$

$\sf f(0,75)=\dfrac{27}{64}+\dfrac{45}{16}+2$

$\sf f(0,75)=\dfrac{27+180+128}{64}$

$\sf f(0,75)=\dfrac{335}{64}$