<ol class="abc-list"><li><div class="list-item-paragraph">The function $\mathrm{f}$ is given by $\mathrm{f}:x\mapsto 1+\sqrt{x}$, for&nbsp;$x\in \mathbb{R}$, $x\ge 0$.</div><ol class="i-list"><li><div class="list-item-paragraph">Find ${{\mathrm{f}}^{-1}}\left( x \right)$ and state the domain of ${{\mathrm{f}}^{-1}}$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Show that if&nbsp;$\mathrm{ff}\left( x \right)=x$ then&nbsp;${{x}^{3}}-4{{x}^{2}}+4x-1=0$. Hence find the value of $x$ for which $\mathrm{ff}\left( x \right)=x$. Explain why this value of $x$ satisfies the equation $\mathrm{f}\left( x \right)={{\mathrm{f}}^{-1}}\left( x \right)$.<span class="inline-marks">[5]</span></div></li></ol></li><li><div class="list-item-paragraph">The function $\mathrm{g}$, with domain the set of non-negative integers, is given by</div><div class="list-item-paragraph" style="text-align: center">$g\left( n \right)=\left\{ \begin{matrix}
1\mathrm{for}\,n=0, \\
2+\mathrm{g}\left( \frac{1}{2}n \right)\mathrm{for \; }n\mathrm{ \; even,} \\
1+\mathrm{g}\left( n-1 \right)\mathrm{for \; }n\mathrm{ \; odd}. \\
\end{matrix} \right.$</div><ol class="i-list"><li><div class="list-item-paragraph">Find $\mathrm{g}\left( 4 \right)$, $\mathrm{g}\left( 7 \right)$and $\mathrm{g}\left( 12 \right)$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Does $\mathrm{g}$ have an inverse? Justify your answer.<span class="inline-marks">[2]</span></div></li></ol></li></ol>

The function $\mathrm{f}$ is given by $\mathrm{f}:x\mapsto 1+\sqrt{x}$, for $x\in \mathbb{R}$, $x\ge 0$.

The function $\mathrm{g}$, with domain the set of non-negative integers, is given by

$g\left( n \right)=\left\{ \begin{matrix}
1\mathrm{for}\,n=0, \\
2+\mathrm{g}\left( \frac{1}{2}n \right)\mathrm{for \; }n\mathrm{ \; even,} \\
1+\mathrm{g}\left( n-1 \right)\mathrm{for \; }n\mathrm{ \; odd}. \\
\end{matrix} \right.$

Functions

International Baccalaureate Higher Level Mathematics

IB HL Math

N2016 P1 Q10

Let $y = 1 + \sqrt{x}$
$y - 1 = \sqrt{x}$
$x = (y - 1)^2$

$\therefore f^{-1}(x) = (x - 1)^2$

$D_{f^{-1}} = R_f = [1, \infty)$

**Method 1:**
$ff(x) = x$
$f^{-1}ff(x) = f^{-1}(x)$
$f(x) = f^{-1}(x)$
$1 + \sqrt{x} = (x - 1)^2$
$1 + \sqrt{x} = x^2 - 2x + 1$
$\sqrt{x} = x^2 - 2x$
$x = (x^2 - 2x)^2$
$x = x^4 - 4x^3 + 4x^2$

Dividing by $x$:
$1 = x^3 - 4x^2 + 4x$
\[x^3 - 4x^2 + 4x - 1 = 0\]

From GC, $x = 0.38196, 1, 2.6180$

Reject $x = 0.38196$ and $x = 1$ as $R_{ff} = [2, \infty)$

$f(1) = 1 + \sqrt{1} = 2$, so $R_{ff} = [2, \infty)$

$\therefore x = 2.62$ (3 s.f.)

Since $ff(x) = x \Rightarrow f^{-1}ff(x) = f^{-1}(x) \Rightarrow f(x) = f^{-1}(x)$

$\therefore$ This value satisfies the equation $f(x) = f^{-1}(x)$.

$g(0) = 1$
$g(1) = 1 + g(0) = 1 + 1 = 2$
$g(2) = 2 + g(1) = 2 + 2 = 4$
$g(3) = 1 + g(2) = 1 + 4 = 5$
$g(4) = 2 + g(2) = 2 + 4 = 6$
$g(5) = 1 + g(4) = 1 + 6 = 7$
$g(6) = 2 + g(3) = 2 + 5 = 7$
$g(7) = 1 + g(6) = 1 + 7 = 8$
$g(12) = 2 + g(6) = 2 + 7 = 9$

$\therefore g(4) = 6, \quad g(7) = 8, \quad g(12) = 9$

$g(5) = g(6) = 7$

The horizontal line $y = 7$ cuts $g$ more than once. Hence, $g$ is not one-one.

Thus, the inverse does not exist.

N2016 P1 Q10

N2016 P1 Q10

Video Solution: