Question

Seja f : R → R definida por f(x) = (2x + 1) × (x - 3)
a) Qual é o valor de $\frac{f(0) + f(1)}{f(-1)}$

b) Quais são as raízes da equação f(x) = -5?​

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Answer 1

a.

$f(x)=(2x+1)(x-3)\\\\f(0)=(2\cdot0+1)(0-3)=1\cdot (-3)=-3\\f(1)=(2 \cdot1+1)(1-3)=(2+1)(-2)=3(-2)=-6\\f(-1)=(2\cdot[-1]+1)(-1-3)=(-1)\cdot(-4)=4\\\\\frac{f(0)+f(1)}{f(-1)}=\frac{-3+(-6)}{4}=\boxed{-\frac{9}{4}}$

b.

$f(x)=(2x+1)(x-3)\\\\(2x+1)(x-3)=-5\\2x^2-6x+x-3+5=0\\2x^2-5x+2=0\\\Delta = (-5)^2 - 4(2)(2)\\\Delta = 25 -4(4)\\\Delta = 25 - 16\\\Delta = 9\\\\x = \frac{-(-5)\pm\sqrt9}{2 \cdot 2}=\frac{5\pm3}{4}\\\\x' = \frac{5+3}{4}=\frac84=\boxed{2}\\\\x" = \frac{5-3}{4}=\frac24=\boxed{\frac12 = 0,5}$