2016 H2 Maths Paper 1, Question 10

Let $y = f(x) = 1 + \sqrt{x}$, for $x \geq 0$. Then

Since $x \geq 0$, this requires $y \geq 1$. Interchanging $x$ and $y$,

The domain of $f^{- 1}$ is the range of $f$. Since $\sqrt{x} \geq 0$ for $x \geq 0$, $f(x) \geq 1$, so the range of $f$ is $x \geq 1$.

Since $f(x) = 1 + \sqrt{x}$,

Setting $ff(x) = x$:

Squaring both sides,

Squaring again,

Note: $x = 0$ does not itself satisfy $ff(x) = x$, since $f(0) = 1$ and $ff(0) = f(1) = 2 \neq 0$; it is an extraneous factor introduced by squaring. Any genuine solution therefore has $x \neq 0$, so it must satisfy the remaining factor:

By inspection, $x = 1$ is a root of this cubic, so it factorises as

Solving $x^{2} - 3x + 1 = 0$ by the quadratic formula gives $x = \frac{3 \pm \sqrt{5}}{2}$.

Note that $x^{2} - 3x + 1 = 0$ is exactly the equation obtained by solving $f(x) = x$ directly: $1 + \sqrt{x} = x$ gives $\sqrt{x} = x - 1$, and squaring gives $x = (x - 1)^{2}$, i.e. $x^{2} - 3x + 1 = 0$ (valid since this needs $x - 1 \geq 0$, i.e. $x \geq 1$). Of the two roots, only $x = \frac{3 + \sqrt{5}}{2}\mkern6mu ( \approx 2.618)$ satisfies $x \geq 1$; the other, $x = \frac{3 - \sqrt{5}}{2}\mkern6mu ( \approx 0.382)$, is rejected.

So $x = \frac{3 + \sqrt{5}}{2}$ satisfies $f(x) = x$.

Since $f(x) = x$, applying $f^{- 1}$ to both sides gives $f^{- 1}\left( f(x) \right) = f^{- 1}(x)$, i.e. $x = f^{- 1}(x)$. Combined with $f(x) = x$, this gives $f(x) = x = f^{- 1}(x)$.

Note: the other candidate, $x = 1$, is rejected, since $f(1) = 2$ but $f^{- 1}(1) = 0$, so $f(1) \neq f^{- 1}(1)$; it is the extraneous root that entered when we squared to derive the cubic (indeed $ff(1) = f(2) = 1 + \sqrt{2} \neq 1$, so $x = 1$ does not even satisfy $ff(x) = x$).

$g$ does not have an inverse.

Explanation: an inverse exists only if $g$ is one-one. But $g(7) = 8$ and $g(8) = 2 + g(4) = 2 + 6 = 8$, so two different elements of the domain, $n = 7$ and $n = 8$, give the same value $g(n) = 8$. Since $g$ is not one-one, $g^{- 1}$ does not exist.