if f(x) = x2+4x+4, then the }domain of y= 1/ f(x) is
a. {x cannot equal -2, 2, xER}
b. {x cannot equal to - 2, xER}
c. {x cannot equal to 2, xER}
d. { xER}
6 years ago
First, determine when f(x)=0.
$f\left(x\right)=x^2+4x+4$ƒ (x)=x2+4x+4
$0=x^2+4x+4$0=x2+4x+4
$0=\left(x+2\right)\left(x+2\right)$0=(x+2)(x+2) or $0=\left(x+2\right)^2$0=(x+2)2 Isolate x
$0=x+2$0=x+2
$x=-2$x=−2
Since the f(x) = 0 when x=-2, then in $y=\frac{1}{x^2+4x+4}$y=1x2+4x+4 , x cannot equal -2 (cannot divide by zero)
Therefore, domain is {x cannot equal to - 2, XER}
totally agree with Andrew T.
6 years ago
Answered By Andrew T
First, determine when f(x)=0.
$f\left(x\right)=x^2+4x+4$ƒ (x)=x2+4x+4
$0=x^2+4x+4$0=x2+4x+4
$0=\left(x+2\right)\left(x+2\right)$0=(x+2)(x+2) or $0=\left(x+2\right)^2$0=(x+2)2 Isolate x
$0=x+2$0=x+2
$x=-2$x=−2
Since the f(x) = 0 when x=-2, then in $y=\frac{1}{x^2+4x+4}$y=1x2+4x+4 , x cannot equal -2 (cannot divide by zero)
Therefore, domain is {x cannot equal to - 2, XER}
6 years ago
Answered By Xuezhong J
totally agree with Andrew T.