Question

1. Seja a função f :ℜ→ℜ definida por

f (x) = x² −10x +8. Calcular os valores reais

de x para que se tenha f (x) = −1, ou seja,

imagem −1 pela função f dada.

2. Dada a função f :ℜ→ℜ definida por

f (x) = ax + b , com a,b∈ℜ, calcular a e b ,

sabendo que f (1) = 4 e f (−1) = −2.

Answer 1

$\boxed{\begin{array}{l}\tt 1)\\\sf x^2-10x+8=-1\\\sf x^2-10x+25+8=25-1\\\sf(x-5)^2+8=24\\\sf\\\sf(x-5)^2=24-8\\\sf (x-5)^2=16\\\sf (x-5)=\pm\sqrt{16}\\\sf (x-5)=\pm4\\\sf x-5=4\\\sf x=4+5\\\sf x=9\\\sf x-5=-4\\\sf x=5-4\\\sf x=1\end{array}}$

$\boxed{\begin{array}{l}\tt 2)\\\sf o~gr\acute afico~passa~pelos~ A(1,4)~e~B(-1,-2)\\\underline{\rm c\acute alculo~do~coeficiente~angular\!:}\\\sf a=\dfrac{y_B-y_A}{x_B-x_A}\\\sf y=\dfrac{-2-4}{-1-1}\\\sf a=\dfrac{-6}{-2}=3\\\sf f(x)=ax+b\\\sf f(1)=3\cdot1+b\\\sf 3+b=4\\\sf b=4-3\\\sf b=1\end{array}}$