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secondary 3 | A Maths
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Please help with the circled parts
Date Posted: 4 years ago
Views: 172
f(x) = ax + b
For the inverse,
x = af-¹(x) + b
For f-¹(7) = -1,
7 = a(-1) + b
7 = -a + b
b = 7 + a
Since 3a - b = 3 from i),
3a - (7 + a) = 3
2a - 7 = 3
2a = 10
a = 5
So b = 7 + 5 = 12
For the inverse,
x = af-¹(x) + b
For f-¹(7) = -1,
7 = a(-1) + b
7 = -a + b
b = 7 + a
Since 3a - b = 3 from i),
3a - (7 + a) = 3
2a - 7 = 3
2a = 10
a = 5
So b = 7 + 5 = 12
h(x) = (4x + 5)/(x - 1)
For the inverse,
x = (4h-¹(x) + 5)/(h-¹(x) - 1)
xh-¹(x) - x = 4h-¹(x) + 5
(x - 4) h-¹(x) = x + 5
h-¹(x) = (x + 5)/(x - 4)
When 2h-¹(2x) = x,
2(2x + 5)/(2x - 4) = x
2(2x + 5) = x(2x - 4)
4x + 10 = 2x² - 4x
2x² - 8x - 10 = 0
x² - 4x - 5 = 0
(x - 5)(x + 1) = 0
x = 5 or x = -1
For the inverse,
x = (4h-¹(x) + 5)/(h-¹(x) - 1)
xh-¹(x) - x = 4h-¹(x) + 5
(x - 4) h-¹(x) = x + 5
h-¹(x) = (x + 5)/(x - 4)
When 2h-¹(2x) = x,
2(2x + 5)/(2x - 4) = x
2(2x + 5) = x(2x - 4)
4x + 10 = 2x² - 4x
2x² - 8x - 10 = 0
x² - 4x - 5 = 0
(x - 5)(x + 1) = 0
x = 5 or x = -1
Thanks
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Thanks J!
Date Posted:
4 years ago
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