3.2 Inverse of Quadratic Function
Quadratic functions are typically not one-to-one over their entire domain. To find an inverse, restrict the domain to a portion where the function is one-to-one. Commonly, restrict to x ≥ h x \ge h x ≥ h x ≤ h x \le h x ≤ h h h h x x x
Example 4
Find the inverse of f ( x ) = x 2 + 4 x + 3 f(x) = x^2 + 4x + 3 f ( x ) = x 2 + 4 x + 3 x ≥ − 2 x \ge -2 x ≥ − 2
Solution
Let y = x 2 + 4 x + 3 y = x^2 + 4x + 3 y = x 2 + 4 x + 3 x ≥ − 2 x \ge -2 x ≥ − 2 x x x y y y
x = y 2 + 4 y + 3 ⟹ y 2 + 4 y + ( 3 − x ) = 0 x = y^2 + 4y + 3 \implies y^2 + 4y + (3 - x) = 0 x = y 2 + 4 y + 3 ⟹ y 2 + 4 y + ( 3 − x ) = 0 y = − 4 ± 16 − 4 ( 3 − x ) 2 = − 4 ± 4 + 4 x 2 = − 4 ± 2 1 + x 2 = − 2 ± 1 + x \begin{align*}
y
&= \dfrac{-4 \pm \sqrt{16 - 4(3 - x)}}{2}
= \dfrac{-4 \pm \sqrt{4 + 4x}}{2}
= \dfrac{-4 \pm 2\sqrt{1 + x}}{2}
= -2 \pm \sqrt{1 + x}
\end{align*} y = 2 − 4 ± 16 − 4 ( 3 − x ) = 2 − 4 ± 4 + 4 x = 2 − 4 ± 2 1 + x = − 2 ± 1 + x So f − 1 ( x ) = − 2 ± 1 + x f^{-1}(x) = -2 \pm \sqrt{1 + x} f − 1 ( x ) = − 2 ± 1 + x f f f
Domain of f = [ − 2 , ∞ ) f = [-2, \infty) f = [ − 2 , ∞ )
f ( x ) = ( x + 2 ) 2 − 1 f(x) = (x + 2)^2 - 1 f ( x ) = ( x + 2 ) 2 − 1 Minimum value is − 1 -1 − 1 f = [ − 1 , ∞ ) f = [-1, \infty) f = [ − 1 , ∞ )
Thus domain of f − 1 = [ − 1 , ∞ ) f^{-1} = [-1, \infty) f − 1 = [ − 1 , ∞ ) f − 1 = [ − 2 , ∞ ) f^{-1} = [-2, \infty) f − 1 = [ − 2 , ∞ )
Check at x = 0 x = 0 x = 0
− 2 + 1 = − 1 , − 2 − 1 = − 3 -2 + \sqrt{1} = -1, \quad -2 - \sqrt{1} = -3 − 2 + 1 = − 1 , − 2 − 1 = − 3 Only − 1 -1 − 1 f − 1 f^{-1} f − 1
Hence
f − 1 ( x ) = − 2 + 1 + x f^{-1}(x) = -2 + \sqrt{1 + x} f − 1 ( x ) = − 2 + 1 + x 