Precalculus
posted by Elli .
Would someone kindly verify these answers~
Thank you!

1.where does the reciprocal function of f(x)=3x increase?
a){XERx>3}
b){XERx<3}
c){XERx≠3}
d){XER}
Answer: C

2. What is the domain of the reciprocal function of f(x)=5x+1?
a) {XERx≥1/5}
b) {XERx≤1/5}
c) {XERx≠1/5
d) {XER}
Answer: C

3. State the range of f(x)= 1/3x+4
a) [YERy>4/3]
b) [YERy≠4/3]
c) [YERy>0]
d) [YERy≠0]
Answer B


1. correct
2. correct
3. did you mean fx) = 1/(3x+4) ?
if so, then the range is d) 
For 3, its the reciprocal of 3x+4
There is no brackets.
3x+4=0
3x=4
x≠4/3
So b? 
@ Luckylin and Reiny, yes, no brackets

If there are no brackets, then the equation is simply the straight line
y = (1/3)x + 4
the range of that line is y ∊R, which is none of the choices, so you must be mistaken
I bet you have a numerator of 1 and a denominator of 3x+4
which has to be written as 1/(3x+4) and I stand by my answer of d)
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