Question

URGENTE!!! 30pt
1 Sejam as funções f(x) = x - 1 e g(x) = 2x^2 - 5x + 3. Determine f(g(x)) e g(f(x)).
2 Seja a função f(x) = 5x + k. Se f(f(2)) = 8. calcule f(3).

Answer 1

$\Large\boxed{\begin{array}{l}\rm 1)\\\sf f(x)=x-1~~~g(x)=2x^2-5x+3\\\sf f(g(x))=g(x)-1\\\sf f(g(x))=2x^2-5x+3-1\\\huge\boxed{\boxed{\boxed{\boxed{\sf f(g(x))=2x^2-5x+2}}}} \\\sf g(f(x))= 2f(x)^2-5f(x)+3\\\sf g(f(x))=2\cdot(x-1)^2-5\cdot(x-1)+3\\\sf g(f(x))=2\cdot(x^2-2x+1)-5x+5+3\\\sf g(f(x))=2x^2-4x+2-5x+5+3\\\huge\boxed{\boxed{\boxed{\boxed{\sf g(f(x))=2x^2-9x+10}}}}\end{array}}$

$\Large\boxed{\begin{array}{l}\rm 2)\\\sf f(x)=5x+k\\\sf f(f(x))=5f(x)+k\\\sf f(f(x))=5\cdot(5x+k)+k\\\sf f(f(x))=25x+5k+k\\\sf f(f(x))=25x+6k\\\sf f(f(2))=25\cdot2+6k\\\sf f(f(2))=50+6k\\\sf 50+6k=8\\\sf 6k=8-50\\\sf 6k=-42\\\sf k=-\dfrac{42}{6}\\\\\sf k=-7\\\sf f(x)=5x-7\\\sf f(3)=5\cdot3-7\\\sf f(3)=15-7\\\huge\boxed{\boxed{\boxed{\boxed{\sf f(3)=8}}}}\end{array}}$