<p>The funtion $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto {{(x-4)}^{2}}+1$ for&nbsp;$x\in \mathbb{R},x&gt;4$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Sketch the graph of $y=\mathrm{f}(x)$. Your sketch should indicate the position of the graph in relation to the origin.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Find ${ {\mathrm{f}}^{-1}}(x)$, stating the domain of ${{\mathrm{f}}^{-1} }$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">On the same diagram as in part <strong>(i)</strong>, sketch the graph of $y={{\mathrm{f}}^{-1}}(x)$.<span class="inline-marks">[1]</span></div></li><li><div class="list-item-paragraph">Write down the equation of the line in which the graph of $y=\mathrm{f(}x\mathrm{)}$ must be reflected in order to obtrain the graph of$y={{\mathrm{f}}^{-1}}(x)$, and hence find the exact solution of the equation $\mathrm{f}(x)={{\mathrm{f}}^{-1}}(x)$.<span class="inline-marks">[5]</span></div></li></ol>

The funtion $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto {{(x-4)}^{2}}+1$ for $x\in \mathbb{R},x>4$.

Sketch the graph of $y=\mathrm{f}(x)$. Your sketch should indicate the position of the graph in relation to the origin.

Find ${ {\mathrm{f}}^{-1}}(x)$, stating the domain of ${{\mathrm{f}}^{-1} }$.

On the same diagram as in part (i), sketch the graph of $y={{\mathrm{f}}^{-1}}(x)$.

On the same diagram as in part **(i)**, sketch the graph of $y={{\mathrm{f}}^{-1}}(x)$.

Write down the equation of the line in which the graph of $y=\mathrm{f(}x\mathrm{)}$ must be reflected in order to obtrain the graph of$y={{\mathrm{f}}^{-1}}(x)$, and hence find the exact solution of the equation $\mathrm{f}(x)={{\mathrm{f}}^{-1}}(x)$.

Modulus Functions

Functions

International Baccalaureate Higher Level Mathematics

IB HL Math

N2008 P2 Q4

$y = f(x) = (x - 4)^2 + 1$ for $x > 4$\n\nVertex at $(4, 1)$ (open circle, as $x > 4$ strictly).\n\nThe curve is the right branch of a parabola, increasing for $x > 4$.

Let $y = (x - 4)^2 + 1$\n\n$(x - 4)^2 = y - 1$\n\n$x - 4 = \pm\sqrt{y - 1}$\n\n$x = 4 + \sqrt{y - 1}$ or $x = 4 - \sqrt{y - 1}$ (reject, as $x > 4$)\n\n$\therefore f^{-1}: x \mapsto 4 + \sqrt{x - 1}, \quad x \in \mathbb{R}, \; x > 1$\n\n$f^{-1}(x) = 4 + \sqrt{x - 1}$, $D_{f^{-1}} = R_f = (1, \infty)$

$y = f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$.

Line of reflection: $y = x$\n\nSince $y = f(x)$, $y = f^{-1}(x)$, and $y = x$ all meet at the same point:\n$f(x) = f^{-1}(x) = x$\n\n$(x - 4)^2 + 1 = x$\n\n$x^2 - 8x + 16 + 1 = x$\n\n$x^2 - 9x + 17 = 0$\n\n$x = \frac{9 \pm \sqrt{(-9)^2 - 4(1)(17)}}{2} = \frac{9 \pm \sqrt{13}}{2}$\n\n$x = \frac{9 + \sqrt{13}}{2}$ or $x = \frac{9 - \sqrt{13}}{2}$ (reject, as $x > 4$)\n\n$\therefore x = \frac{9 + \sqrt{13}}{2}$

N2008 P2 Q4

N2008 P2 Q4

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