Problems 1-8 : Find f-1(x).
Problem 1 :
f(x) = {(2, 3), (3, 5), (4, 7), (5, 9)}
Solution :
f-1(x) = {(3, 2), (5, 3), (7, 4), (9, 5)}
Problem 2 :
Solution :
Replace f(x) by y.
Switch x and y.
Solve for y interms of x.
Replace y by f-1(x).
Problem 3 :
f(x) = 3√x - 5
Solution :
f(x) = 3√x - 5
y = 3√x - 5
x = 3√y - 5
x + 5 = 3√y
(x + 5)3 = (3√y)3
(x + 5)3 = y
f-1(x) = (x + 5)3
Problem 4 :
Solution :
y + 7 = (3 - x)2
y = (3 - x)2 - 7
Problem 5 :
f(x) = 4 - x2
Solution :
f(x) = 4 - x2
y = 4 - x2
x = 4 - y2
y2 = 4 - x
Problem 6 :
f(x) = x2 + 6x + 7
Solution :
f(x) = x2 + 6x + 7
y = x2 + 6x + 7
x = y2 + 6y + 7
x = y2 + 2(y)(3) + 32 - 32 + 7
x = (y + 3)2 - 32 + 7
x = (y + 3)2 - 9 + 7
x = (y + 3)2 - 2
x + 2 = (y + 3)2
(y + 3)2 = x + 2
Problem 7 :
f(x) = log2(x + 5)
Solution :
f(x) = log2(x + 5)
y = log2(x + 5)
x = log2(y + 5)
Convert the above equation to exponential.
2x = y + 5
2x - 5 = y
y = 2x - 5
f-1(x) = 2x - 5
Problem 8 :
f(x) = 10x - 3 + 2
Solution :
f(x) = 10x - 3 + 2
y = 10x - 3 + 2
x = 10y - 3 + 2
x - 2 = 10y - 3
Convert the above equation to logarithmic form.
log10(x - 2) = y - 3
log10(x - 2) + 3 = y
y = log10(x - 2) + 3
f-1(x) = log10(x - 2) + 3
Problem 9 :
The function f(x) = a(-x3 - x + 2) has an inverse function such that f-1(6) = -2. Solve for a.
Solution :
f-1(6) = -2
-2 = f-1(6)
f(-2) = f[f-1(6)]
f(-2) = 6
a[-(-2)3 - (-2) + 2] = 6
a[-(-8) + 2 + 2] = 6
a(8 + 4) = 6
a(12) = 6
12a = 6
Problem 10 :
If the point P(-2, 5) is on the graph of y = f(x), then determine the coordinates of P' on the graph of
y = f-1(x - 3) + 1
Solution :
When P(-2, 5) is on the graph of y = f(x), let P'(a, b) is on the graph of y = f-1(x - 3) + 1.
Then, we have
5 = f(-2) or f(-2) = 5 ----(1) |
b = f-1(a - 3) + 1 b - 1 = f-1(a - 3) f(b - 1) = f[f-1(a - 3)] f(b - 1) = a - 3 ----(2) |
Comparing (1) and (2), we get
b - 1 = -2 b = -1 |
a - 3 = 5 a = 8 |
The coordinates of P' are
(a, b) = (8, -1)
Therefore, if the point P(-2, 5) is on the graph of y = f(x), then the coordinates of P' on the graph of y = f-1(x - 3) + 1 are (8, -1).
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