SOLVING PIECEWISE FUNCTIONS

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

x ≈ -1.414 or 1.414

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

x ≈ 1.414 ∈ 0 ≤ x < 2

Piece 3 :

-e^{2x + 2} + 5 = 1

-e^{2x + 2} = -4

Multiply both sides by -1.

e^{2x + 2} = 4

2x + 2 = ln(4)

2x = ln(4) - 2

x = ½[ln(4) - 2]

x ≈ -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

x ≈ ±2.236 ∉ 0 ≤ x < 2

NO SOLUTION

Piece 3 :

-e^{2x + 2} + 5 = 4

-e^{2x + 2} = -1

Multiply both sides by -1.

e^{2x + 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

x ≈ ±2.828 ∉ 0 ≤ x < 2

NO SOLUTION

Piece 3 :

-e^{2x + 2} + 5 = 7

-e^{2x + 2} = 2

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

NO SOLUTION

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