SOLVING PIECEWISE FUNCTIONS

Given the piece wise function f(x).

Question 1 :

Solve : f(x) = 1.

Answer :

Piece 1 :

x + 2 = 9

x = 7 ∈ x ≥ 2

Piece 2 :

x2 - 1 = 1

x2 = 2

x ±√2

≈ -1.414 or 1.414

x = -1.414 is not in the interval 0 ≤ x < 2.

x  1.414 ∈ ≤ x < 2

Piece 3 :

-e2x + 2 + 5 = 1

-e2x + 2 = -4

Multiply both sides by -1.

e2x + 2 = 4

2x + 2 = ln(4)

2x = ln(4) - 2

x = ½[ln(4) - 2]

≈ -0.307 ∈ x < 0

Question 2 :

Solve : f(x) = 4.

Answer :

Piece 1 :

x + 2 = 36

x = 34 ∈ x ≥ 2

Piece 2 :

x2 - 1 = 4

x2 = 5

x ±√5

±2.236  ≤ x < 2

NO SOLUTION

Piece 3 :

-e2x + 2 + 5 = 4

-e2x + 2 = -1

Multiply both sides by -1.

e2x + 2 = 1

2x + 2 = ln(1)

2x + 2 = 0

2x = -2

x = -1 ∈ x < 0

Question 3 :

Solve : f(x) = 7.

Answer :

Piece 1 :

x + 2 = 81

x = 79 ∈ x ≥ 2

Piece 2 :

x2 - 1 = 7

x2 = 8

x ±√8

±2.828  ≤ x < 2

NO SOLUTION

Piece 3 :

-e2x + 2 + 5 = 7

-e2x + 2 = 2

Soince a negative value  a positive value, the above equation is false.

NO SOLUTION

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