Posted by **Elli** on Thursday, December 22, 2011 at 7:21pm.

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}

Answer: C

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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}

Answer: C

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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]

Answer B

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- Precalculus -
**Reiny**, Thursday, December 22, 2011 at 7:58pm
1. correct

2. correct

3. did you mean fx) = 1/(3x+4) ?

if so, then the range is d)

- Precalculus -
**Luckylin**, Thursday, December 22, 2011 at 8:04pm
For 3, its the reciprocal of 3x+4

There is no brackets.

3x+4=0

3x=-4

x≠-4/3

So b?

- Precalculus -
**Elli**, Thursday, December 22, 2011 at 8:22pm
@ Luckylin and Reiny, yes, no brackets

- Precalculus -
**Reiny**, Thursday, December 22, 2011 at 8:38pm
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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