Worksheet: Simplifying Rational Functions

Practice

In this worksheet, we will practice simplifying rational functions and finding their domains.

Q1:

Simplify the function , and find its domain.

  • A , domain
  • B , domain
  • C , domain
  • D , domain
  • E , domain

Q2:

Simplify the function ๐‘› ( ๐‘ฅ ) = ๐‘ฅ + 1 ๐‘ฅ + 3 ๐‘ฅ + 2 ๏Šจ and find its domain.

  • A ๐‘› ( ๐‘ฅ ) = 1 ๐‘ฅ โˆ’ 2 , domain = โ„ โˆ’ { โˆ’ 1 , 2 }
  • B ๐‘› ( ๐‘ฅ ) = 1 ๐‘ฅ + 2 , domain = โ„ โˆ’ { โˆ’ 2 }
  • C ๐‘› ( ๐‘ฅ ) = 1 ๐‘ฅ โˆ’ 2 , domain = โ„ โˆ’ { 2 }
  • D ๐‘› ( ๐‘ฅ ) = 1 ๐‘ฅ + 2 , domain = โ„ โˆ’ { โˆ’ 1 , โˆ’ 2 }
  • E ๐‘› ( ๐‘ฅ ) = ๐‘ฅ + 1 ( ๐‘ฅ โˆ’ 1 ) ( ๐‘ฅ โˆ’ 2 ) , domain = โ„ โˆ’ { 1 , 2 }

Q3:

Given the function ๐‘“ ( ๐‘ฅ ) = 7 4 ๐‘ฅ โˆ’ 8 1 + 1 9 ๐‘ฅ โˆ’ 2 ๐‘ฅ ๏Šจ ๏Šจ , evaluate ๐‘“ ( 3 ) .

  • A ๐‘“ ( 3 ) = โˆ’ 2 9
  • B ๐‘“ ( 3 ) = 3 4 4 5
  • C ๐‘“ ( 3 ) = 8 4 5
  • D ๐‘“ ( 3 ) = โˆ’ 2 4 5

Q4:

Simplify the function ๐‘“ ( ๐‘ฅ ) = 7 ๐‘ฅ + 4 3 ๐‘ฅ + 6 7 ๐‘ฅ + 5 0 ๐‘ฅ + 7 2 2 , and find its domain.

  • A ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 6 ๐‘ฅ โˆ’ 7 , domain = โ„ โˆ’ ๏ฌ โˆ’ 1 7 , โˆ’ 7 ๏ธ
  • B ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ + 6 ๐‘ฅ + 7 , domain = โ„ โˆ’ { โˆ’ 7 }
  • C ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 6 ๐‘ฅ โˆ’ 7 , domain = โ„ โˆ’ { โˆ’ 7 }
  • D ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ + 6 ๐‘ฅ + 7 , domain = โ„ โˆ’ ๏ฌ โˆ’ 1 7 , โˆ’ 7 ๏ธ
  • E ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 6 ๐‘ฅ + 7 , domain = โ„ โˆ’ ๏ฌ โˆ’ 1 7 , โˆ’ 7 ๏ธ

Q5:

Given that ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 5 ๐‘ฅ + 5 ๏Šง and ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 5 ๐‘ฅ ๐‘ฅ + 5 ๐‘ฅ ๏Šจ ๏Šจ ๏Šจ , find the largest set on which the functions ๐‘› ๏Šง and ๐‘› ๏Šจ are equal.

  • A โ„ โˆ’ { โˆ’ 5 }
  • B โ„ โˆ’ { 0 , 5 }
  • C โ„ โˆ’ { 5 }
  • D โ„ โˆ’ { โˆ’ 5 , 0 }
  • E โ„ โˆ’ { โˆ’ 5 , 0 , 5 }

Q6:

Given the functions ๐‘› ( ๐‘ฅ ) = ๐‘ฅ ๐‘ฅ โˆ’ 1 0 ๐‘ฅ ๏Šง ๏Šจ and ๐‘› ( ๐‘ฅ ) = 1 ๐‘ฅ โˆ’ 1 0 ๏Šจ , what is the set of values on which ๐‘› = ๐‘› ๏Šง ๏Šจ ?

  • A โ„ โˆ’ { 1 0 }
  • B โ„ โˆ’ { โˆ’ 1 0 , 0 }
  • C โ„ โˆ’ { 0 }
  • D โ„ โˆ’ { 0 , 1 0 }
  • E { 0 }

Q7:

Which of the following statements describes when two functions ๐‘› 1 and ๐‘› 2 are equal?

  • A ๐‘› ( ๐‘ฅ ) = ๐‘› ( ๐‘ฅ ) 1 2
  • B the domain of ๐‘› = 1 the domain of ๐‘› 2
  • C the domain of ๐‘› = 1 the domain of ๐‘› 2 and ๐‘› ( ๐‘ฅ ) โ‰  ๐‘› ( ๐‘ฅ ) 1 2
  • Dthe domain of ๐‘› = 1 the domain of ๐‘› 2 and ๐‘› ( ๐‘ฅ ) = ๐‘› ( ๐‘ฅ ) 1 2 for each ๐‘ฅ in the common domain

Q8:

Simplify the function ๐‘› ( ๐‘ฅ ) = ๐‘ฅ + 1 ( ๐‘ฅ + 1 ) ( ๐‘ฅ โˆ’ 3 ) ๏Šฉ and find its domain.

  • A ๐‘› ( ๐‘ฅ ) = ( ๐‘ฅ + 1 ) ๐‘ฅ โˆ’ 3 ๏Šจ , domain = โ„ โˆ’ { โˆ’ 1 , 3 }
  • B ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ ๐‘ฅ + 1 ๐‘ฅ โˆ’ 3 ๏Šจ , domain = โ„ โˆ’ { 3 }
  • C ๐‘› ( ๐‘ฅ ) = ( ๐‘ฅ + 1 ) ๐‘ฅ โˆ’ 3 ๏Šจ , domain = โ„ โˆ’ { 3 }
  • D ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ ๐‘ฅ + 1 ๐‘ฅ โˆ’ 3 ๏Šจ , domain = โ„ โˆ’ { โˆ’ 1 , 3 }
  • E ๐‘› ( ๐‘ฅ ) = ๐‘ฅ + ๐‘ฅ + 1 ๐‘ฅ โˆ’ 3 ๏Šจ , domain = โ„ โˆ’ { โˆ’ 1 , 3 }

Q9:

Simplify the function ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 8 1 ๐‘ฅ + 7 2 9 ๏Šจ ๏Šฉ and find its domain.

  • A ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ + 9 ๐‘ฅ + 9 ๐‘ฅ + 8 1 ๏Šจ , domain = โ„ โˆ’ { โˆ’ 9 }
  • B ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 9 ๐‘ฅ โˆ’ 9 ๐‘ฅ + 8 1 ๏Šจ , domain = โ„
  • C ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ + 9 ๐‘ฅ + 9 ๐‘ฅ + 8 1 ๏Šจ , domain = โ„
  • D ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 9 ๐‘ฅ โˆ’ 9 ๐‘ฅ + 8 1 ๏Šจ , domain = โ„ โˆ’ { โˆ’ 9 }
  • E ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 9 ๐‘ฅ โˆ’ 9 ๐‘ฅ + 8 1 ๏Šจ , domain = โ„ โˆ’ { 9 }

Q10:

Simplify the function ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 1 2 5 ๐‘ฅ + 5 ๐‘ฅ + 2 5 ๏Šฌ ๏Šช ๏Šจ , and find its domain.

  • A ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 5 ๏Šจ , domain = โ„
  • B ๐‘› ( ๐‘ฅ ) = ๏€ป ๐‘ฅ โˆ’ โˆš 5 ๏‡ ๏€ป ๐‘ฅ + โˆš 5 ๏‡ , domain = โ„ โˆ’ { 5 }
  • C ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 5 ๏Šจ , domain = โ„ โˆ’ { โˆ’ 5 }
  • D ๐‘› ( ๐‘ฅ ) = ๏€ป ๐‘ฅ โˆ’ โˆš 5 ๏‡ ๏€ป ๐‘ฅ + โˆš 5 ๏‡ , domain = โ„
  • E ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 5 ๏Šฉ , domain = โ„

Q11:

Simplify the function ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ ๏Šฉ + ๐‘ฅ ๏Šจ โˆ’ 8 0 ๐‘ฅ โˆ’ 4 , and find its domain.

  • A ๐‘ฅ ๏Šจ โˆ’ 3 ๐‘ฅ + 2 0 , domain = โ„ โˆ’ { 4 }
  • B ๐‘ฅ ๏Šจ + 5 ๐‘ฅ + 2 0 , domain = โ„
  • C ๐‘ฅ ๏Šจ โˆ’ 3 ๐‘ฅ + 2 0 , domain = โ„
  • D ๐‘ฅ ๏Šจ + 5 ๐‘ฅ + 2 0 , domain = โ„ โˆ’ { 4 }
  • E ๐‘ฅ ๏Šจ + 4 ๐‘ฅ + 2 0 , domain = โ„ โˆ’ { 4 }

Q12:

Simplify the function ๐‘› ( ๐‘ฅ ) = ๐‘ฅ + ๐‘ฅ โˆ’ 2 0 ๐‘ฅ + 5 ๐‘ฅ โˆ’ 1 6 ๐‘ฅ โˆ’ 8 0 ๏Šจ ๏Šฉ ๏Šจ , and find its domain.

  • A ๐‘› ( ๐‘ฅ ) = 1 ๐‘ฅ โˆ’ 4 , domain = โ„ โˆ’ { โˆ’ 5 , โˆ’ 4 , 4 }
  • B ๐‘› ( ๐‘ฅ ) = 1 ๐‘ฅ + 4 , domain = โ„ โˆ’ { 4 }
  • C ๐‘› ( ๐‘ฅ ) = 1 ๐‘ฅ โˆ’ 4 , domain = โ„ โˆ’ { โˆ’ 4 }
  • D ๐‘› ( ๐‘ฅ ) = 1 ๐‘ฅ + 4 , domain = โ„ โˆ’ { โˆ’ 5 , โˆ’ 4 , 4 }
  • E ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 4 ๐‘ฅ + 1 6 ๏Šจ , domain = โ„ โˆ’ { โˆ’ 5 , โˆ’ 4 , 4 }

Q13:

Given that the algebraic fraction ๐‘› ( ๐‘ฅ ) = 8 ๐‘ฅ ( ๐‘ฅ + 4 ) ๐‘ฅ + ๐‘Ž simplifies to ๐‘› ( ๐‘ฅ ) = 8 ๐‘ฅ , what is the value of ๐‘Ž ?

Q14:

Given that ๐‘› ( ๐‘ฅ ) = ๐‘ฅ + 6 4 ๐‘ฅ โˆ’ 1 6 ๏Šง , ๐‘› ( ๐‘ฅ ) = 4 ๐‘ฅ + 2 5 6 ๐‘ฅ โˆ’ 1 6 ๏Šจ , and ๐‘› ( ๐‘ฅ ) = ๐‘› ( ๐‘ฅ ) รท ๐‘› ( ๐‘ฅ ) ๏Šง ๏Šจ , find ๐‘› ( โˆ’ 4 ) if possible.

  • A 1 6 4
  • B4
  • C 1 2
  • D 1 4
  • E64

Q15:

Given that ๐‘› ( ๐‘ฅ ) = ๐‘ฅ + 1 2 ๐‘ฅ + 3 6 ๐‘ฅ โˆ’ ๐‘Ž ๏Šจ ๏Šจ simplifies to ๐‘› ( ๐‘ฅ ) = ๐‘ฅ + 6 ๐‘ฅ โˆ’ 6 , what is the value of ๐‘Ž ?

Q16:

Simplify the function ๐‘“ ( ๐‘ฅ ) = ( ๐‘ฅ + 3 ) โˆ’ 3 6 ๐‘ฅ ( ๐‘ฅ โˆ’ 3 ) ๏Šจ , and find its domain.

  • A ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 9 ๐‘ฅ , domain = โ„ โˆ’ { 0 , 3 }
  • B ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ + 9 ๐‘ฅ , domain = โ„ โˆ’ { 0 }
  • C ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 9 ๐‘ฅ , domain = โ„ โˆ’ { 0 }
  • D ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ + 9 ๐‘ฅ , domain = โ„ โˆ’ { 0 , 3 }
  • E ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ + 9 ๐‘ฅ , domain = โ„ โˆ’ { 0 , โˆ’ 3 }

Q17:

Given that the functions ๐‘› ( ๐‘ฅ ) = 8 ๐‘ฅ ๐‘ฅ + ๐‘ ๏Šง and ๐‘› ( ๐‘ฅ ) = 8 ๐‘ฅ + ๐‘‘ ๐‘ฅ ๐‘ฅ + ๐‘ ๐‘ฅ + 5 ๐‘ฅ โˆ’ 1 5 ๏Šจ ๏Šฉ ๏Šฉ ๏Šจ are equal, what are the values of ๐‘ and ๐‘‘ ?

  • A ๐‘ = โˆ’ 3 , ๐‘‘ = โˆ’ 4 0
  • B ๐‘ = 3 , ๐‘‘ = 4 0
  • C ๐‘ = โˆ’ 3 , ๐‘‘ = 5
  • D ๐‘ = โˆ’ 3 , ๐‘‘ = 4 0
  • E ๐‘ = 3 , ๐‘‘ = โˆ’ 4 0

Q18:

Which of the following functions are equal?

  • A ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 7 2 9 ๐‘ฅ + 9 ๐‘ฅ + 8 1 ๐‘ฅ ๏Šง ๏Šฉ ๏Šฉ ๏Šจ , ๐‘› ( ๐‘ฅ ) = ( ๐‘ฅ โˆ’ 9 ) ( ๐‘ฅ + 6 3 ) ๐‘ฅ + 6 3 ๐‘ฅ ๏Šจ ๏Šจ
  • B ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 7 2 9 ๐‘ฅ + 9 ๐‘ฅ + 8 1 ๐‘ฅ ๏Šง ๏Šฉ ๏Šฉ ๏Šจ , ๐‘› ( ๐‘ฅ ) = ( ๐‘ฅ โˆ’ 9 ) ๏€น ๐‘ฅ โˆ’ 6 3 ๏… ๐‘ฅ โˆ’ 6 3 ๐‘ฅ ๏Šจ ๏Šจ ๏Šฉ
  • C ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 7 2 9 ๐‘ฅ + 9 ๐‘ฅ + 8 1 ๐‘ฅ ๏Šง ๏Šฉ ๏Šฉ ๏Šจ , ๐‘› ( ๐‘ฅ ) = ( ๐‘ฅ โˆ’ 9 ) ( ๐‘ฅ + 6 3 ) ๐‘ฅ + 6 3 ๐‘ฅ ๏Šจ ๏Šฉ
  • D ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 7 2 9 ๐‘ฅ + 9 ๐‘ฅ + 8 1 ๐‘ฅ ๏Šง ๏Šฉ ๏Šฉ ๏Šจ , ๐‘› ( ๐‘ฅ ) = ( ๐‘ฅ โˆ’ 9 ) ๏€น ๐‘ฅ + 6 3 ๏… ๐‘ฅ + 6 3 ๐‘ฅ ๏Šจ ๏Šจ ๏Šฉ
  • E ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 7 2 9 ๐‘ฅ + 9 ๐‘ฅ + 8 1 ๐‘ฅ ๏Šง ๏Šฉ ๏Šฉ ๏Šจ , ๐‘› ( ๐‘ฅ ) = ( ๐‘ฅ โˆ’ 9 ) ( ๐‘ฅ โˆ’ 6 3 ) ๐‘ฅ โˆ’ 6 3 ๐‘ฅ ๏Šจ ๏Šฉ

Q19:

Simplify the function ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ + 4 ๐‘ฅ + 4 5 ๐‘ฅ โˆ’ 2 0 ๐‘ฅ ๏Šจ ๏Šฉ and find its domain.

  • A ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 2 5 ๐‘ฅ ( ๐‘ฅ + 2 ) , domain = โ„ โˆ’ { 0 , โˆ’ 2 , 2 }
  • B ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ + 2 5 ๐‘ฅ ( ๐‘ฅ โˆ’ 2 ) , domain = โ„ โˆ’ { 0 , 2 }
  • C ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 2 5 ๐‘ฅ ( ๐‘ฅ + 2 ) , domain = โ„ โˆ’ { 0 , โˆ’ 2 }
  • D ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ + 2 5 ๐‘ฅ ( ๐‘ฅ โˆ’ 2 ) , domain = โ„ โˆ’ { 0 , โˆ’ 2 , 2 }
  • E ๐‘“ ( ๐‘ฅ ) = ( ๐‘ฅ + 2 ) 5 ๐‘ฅ ( ๐‘ฅ โˆ’ 4 ) ๏Šจ ๏Šจ , domain = โ„ โˆ’ { 0 , โˆ’ 2 , 2 }

Q20:

Which of the following functions are equal?

  • A ,
  • B ,
  • C ,
  • D ,
  • E ,

Q21:

Given that ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ ๐‘Ž ๐‘ฅ โˆ’ 3 2 ๐‘ฅ + ๐‘ฅ โˆ’ 7 2 ๏Šจ ๏Šจ , and the multiplicative inverse of ๐‘› is ๐‘ฅ + 9 ๐‘ฅ + 4 , what is the value of ๐‘Ž ?

Q22:

Given the functions ๐‘ ( ๐‘ฅ ) = 3 ๐‘ฅ โˆ’ 3 0 ๐‘ฅ ( ๐‘ฅ + 1 0 ) ( ๐‘ฅ โˆ’ 1 0 ) 2 and ๐‘ž ( ๐‘ฅ ) = 3 ๐‘ฅ ๐‘ฅ + 1 0 , what is the set of values on which ๐‘ = ๐‘ž ?

  • A โ„ โˆ’ { 1 0 }
  • B โ„ โˆ’ { โˆ’ 1 0 }
  • C โ„ โˆ’ { 0 , 1 0 }
  • D โ„ โˆ’ { 1 0 , โˆ’ 1 0 }
  • E โ„ โˆ’ { โˆ’ 1 0 , 0 }

Q23:

Given that the multiplicative inverse of the function ๐‘› ( ๐‘ฅ ) = 2 ๐‘ฅ + 1 0 ๐‘ฅ ๐‘ฅ + 1 4 ๐‘ฅ + ๐‘Ž 2 2 is ๐‘ฅ + 9 2 ๐‘ฅ , find the value of ๐‘Ž .

Q24:

Determine the domain of the function ๐‘› ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 6 4 8 ๐‘ฅ + 7 ๐‘ฅ รท 9 ๐‘ฅ โˆ’ 1 1 7 ๐‘ฅ + 3 6 0 6 4 ๐‘ฅ โˆ’ 4 9 ๏Šจ ๏Šจ ๏Šจ ๏Šจ .

  • A โ„ โˆ’ ๏ฌ โˆ’ 7 8 , 0 , 7 8 ๏ธ
  • B โ„ โˆ’ ๏ฌ โˆ’ 7 8 , 7 8 , 5 , 8 ๏ธ
  • C โ„ โˆ’ ๏ฌ โˆ’ 7 8 , 0 , 5 , 8 ๏ธ
  • D โ„ โˆ’ ๏ฌ โˆ’ 7 8 , 0 , 7 8 , 5 , 8 ๏ธ
  • E โ„ โˆ’ { 0 , 5 }

Q25:

Simplify the function ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 4 ๐‘ฅ โˆ’ ๐‘ฅ โˆ’ 2 ๏Šจ ๏Šจ and find its domain.

  • A ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 2 ๐‘ฅ โˆ’ 1 , domain = โ„ โˆ’ { โˆ’ 2 , 1 }
  • B ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ + 2 ๐‘ฅ + 1 , domain = โ„ โˆ’ { โˆ’ 1 }
  • C ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ โˆ’ 2 ๐‘ฅ โˆ’ 1 , domain = โ„ โˆ’ { 1 }
  • D ๐‘“ ( ๐‘ฅ ) = ๐‘ฅ + 2 ๐‘ฅ + 1 , domain = โ„ โˆ’ { 2 , โˆ’ 1 }
  • E ๐‘“ ( ๐‘ฅ ) = โˆ’ 4 โˆ’ ๐‘ฅ โˆ’ 2 , domain = โ„ โˆ’ { 2 , โˆ’ 1 }

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