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Sagot :
(x + 1) y = 1
Solution:
Step 1: Isolate y to -re-write the expression as function, y = f(x)
[tex] \frac{(x+1)y}{x+1} = \frac{1}{x+1} [/tex]
f(x) = [tex] \frac{1}{x+1} [/tex]
Step 2: Solve for domain.
x + 1 = 0
x - 1 = 0 - 1
x = -1
Since x + 1 is the denominator of [tex] \frac{1}{x+1} [/tex], when -1 is substituted to denominator x + 1 ⇒ (-1) + 1 = 0.
Note that in any rational expression, denominator can not be 0, because it will render the expression 'undefined".
Therefore:
Domain = {x/x ≠ -1)
= x < -1, x > -1
= {-2, -3, -4,...} and {0, 1, 2, 3, ...}
Interval Notation (⁻∞, -1) U (-1, ⁺∞)
Step 3: Solve for range
Remember that the range is the combined domain of its inverse functions.
f⁻¹(x) = -1 + 1/x
f⁻¹(x) = [tex] \frac{1-x}{x} [/tex]
Find the value of denominator x:
x = 0
Again, if the denominator is equal to 0, the rational expression is undefined.
Therefore:
Range = {y/y ≠0}
= y < 0, y > 0
= {-1, -2, -3, ...} and {1, 2, 3, ...}
Interval Notation: (⁻∞, 0) U (0, ⁺∞)
Solution:
Step 1: Isolate y to -re-write the expression as function, y = f(x)
[tex] \frac{(x+1)y}{x+1} = \frac{1}{x+1} [/tex]
f(x) = [tex] \frac{1}{x+1} [/tex]
Step 2: Solve for domain.
x + 1 = 0
x - 1 = 0 - 1
x = -1
Since x + 1 is the denominator of [tex] \frac{1}{x+1} [/tex], when -1 is substituted to denominator x + 1 ⇒ (-1) + 1 = 0.
Note that in any rational expression, denominator can not be 0, because it will render the expression 'undefined".
Therefore:
Domain = {x/x ≠ -1)
= x < -1, x > -1
= {-2, -3, -4,...} and {0, 1, 2, 3, ...}
Interval Notation (⁻∞, -1) U (-1, ⁺∞)
Step 3: Solve for range
Remember that the range is the combined domain of its inverse functions.
f⁻¹(x) = -1 + 1/x
f⁻¹(x) = [tex] \frac{1-x}{x} [/tex]
Find the value of denominator x:
x = 0
Again, if the denominator is equal to 0, the rational expression is undefined.
Therefore:
Range = {y/y ≠0}
= y < 0, y > 0
= {-1, -2, -3, ...} and {1, 2, 3, ...}
Interval Notation: (⁻∞, 0) U (0, ⁺∞)
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