Question

Sejam as funções F e G reais definidas por F(x)= x-1 e G(x)= x²-3. Determine:
a) F(g(x))
b) G(f(x))
c) G(g(x))
d) F(g(x)) = 0
e) G(f(x)) = 1

Answer 1

a) $f(x)=x-1 \\\\ f(g(x))=g(x)-1 \\\\ f(g(x))=x^{2}-3-1 \\\\ \boxed{\boxed{f(g(x))=x^{2}-4}}$


b) $g(x) = x^{2}-3 \\\\ g(f(x))=f(x)^{2}-3 \\\\ g(f(x))=(x-1)^{2}-3 \\\\ g(f(x))=x^{2}-2x+1-3 \\\\ \boxed{\boxed{g(f(x))=x^{2}-2x-2}}$


c) $g(x)=x^{2}-3 \\\\ g(g(x))=(x^{2}-3)^{2}-3 \\\\ g(g(x)) = x^{4}-6x^{2}+9-3 \\\\ \boxed{\boxed{g(g(x)) = x^{4}-6x^{2}+6}}$


d) O mesmo resultado da A igualado a zero.

$f(g(x))=x^{2}-4 \\\\ x^{2}-4 = 0 \\\\ x^{2} = 4 \\\\ x = \pm \sqrt{4} \\\\ \boxed{\boxed{x = \pm 2 }}$


e) $g(f(x))=x^{2}-2x-2 \\\\ x^{2}-2x-2=1 \\\\ x^{2}-2x-2-1=0 \\\\ x^{2}-2x-3=0 \\\\ \Delta = b^{2}-4 \cdot a \cdot c \\\\ \Delta = (-2)^{2}-4 \cdot (1) \cdot (-3) \\\\ \Delta = 4+12 \\\\ \Delta = 16$


$x = \frac{-b \pm \sqrt{\Delta}}{2a} \\\\ x = \frac{-(-2) \pm \sqrt{16}}{2 \cdot 1} \\\\ x = \frac{2 \pm 4}{2} \\\\\\ x_{1} = \frac{2 + 4}{2} = \frac{6}{2} = \boxed{3} \\\\ x_{2} = \frac{2 - 4}{2} = -\frac{2}{2} = \boxed{-1}$