Post a New Question

Precalculus

posted by on .

Would someone kindly verify these answers~

Thank you!
---------------------------------------
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

----------------------------------------

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
--------------------------------------

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
---------------------------------------

  • 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)

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question