Question

dada a função f: R ⇒R (ou seja o domínio e o contra domínio são os números reais) definida por f(x)= x ao quadrado-5x+6 , calcule:
a) f(-1); f(-4) e f(-5)
b) o valor de x cujo imagem igual a 2.

Answer 1

$f(x)= x^{2} -5x+6 \\ \\ a) \\ f(-1) =(-1)^{2} -5(-1)+6=1+5+6=12 \\ f(-4)=(-4)^{2} -5(-4)+6=16+20+6=42 \\ f(-5)=(-5)^{2} -5(-5)+6=25+25+6=56 \\ \\ b) \\ f(x)=x^{2} -5x+6=2 \\ x^{2} -5x+6-2=0==\ \textgreater \ x^{2} -5x+4=0 \\ x_{12}= \frac{-b+/- \sqrt{b^2-4ac} }{2a} = \frac{-(-5)+/- \sqrt{(-5)^2-4(1)(4)} }{2(1)}= \\ =\frac{5+/- \sqrt{25-16} }{2}=\frac{5+/- \sqrt{9} }{2}=\frac{5+/- 3 }{2}==\ \textgreater \ \\ x_{1} =4 \\ x_{2} =1$

Answer 2

f(x)=x2−5x+6a)f(−1)=(−1)2−5(−1)+6=1+5+6=12f(−4)=(−4)2−5(−4)+6=16+20+6=42f(−5)=(−5)2−5(−5)+6=25+25+6=56b)f(x)=x2−5x+6=2x2−5x+6−2=0== > x2−5x+4=0x12​=2a−b+/−b2−4ac​​=2(1)−(−5)+/−(−5)2−4(1)(4)​​==25+/−25−16​​=25+/−9​​=25+/−3​== > x1​=4x2​=1​