# Precalculus

posted by .

Would someone kindly verify these answers~

Thank you!
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1.where does the reciprocal function of f(x)=3-x increase?

a){XER|x>3}
b){XER|x<3}
c){XER|x≠3}
d){XER}

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2. What is the domain of the reciprocal function of f(x)=5x+1?

a) {XER|x≥-1/5}
b) {XER|x≤-1/5}
c) {XER|x≠-1/5
d) {XER}

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3. State the range of f(x)= 1/3x+4

a) [YER|y>-4/3]
b) [YER|y≠-4/3]
c) [YER|y>0]
d) [YER|y≠0]

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• Precalculus -

1. correct
2. correct
3. did you mean fx) = 1/(3x+4) ?
if so, then the range is d)

• Precalculus -

For 3, its the reciprocal of 3x+4

There is no brackets.

3x+4=0
3x=-4
x≠-4/3

So b?

• Precalculus -

@ Luckylin and Reiny, yes, no brackets

• Precalculus -

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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