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David’s problem of the week

by dvdalbertini on May 22, 2011

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David’s Problem of the Week 5/16-5/20
f(x)= | x^2 – 5 | Find the inverse of f(x) and give 5 points of the inverse.
The Solution
- f(x)= | x² – 5 |
- y= | x² – 5 |
- y= ±(x² - 5)
- y= -(x² - 5)
- y= - x² + 5
- 5 - 5
- (-y – 5)(-1) = (- x²)(-1)
- √ (-y + 5)= √(x²)
- √ (-y + 5) = | x |
- √ (-y + 5) = ±x
- ±√(-y + 5) =x
- √ (-x + 5) = y
y = x² - 5 +5 +5 √ (y + 5) = (x²) √ (y + 5) = |x| √ (y + 5) = ±x ±√(y + 5) = x √ (x + 5) = y
The Solution Note: Since the function is absolute value, the inverse must be piecewise in this case √ (x + 5) = y when x > √5 √ (-x + 5) = y when 0<x< √5 f^(-1)(x)=