<p>Functions $\mathrm{f}$ and $\mathrm{g}$ are defined by</p>
<p style="margin-left: 9rem">$\mathrm{f}:x\mapsto {{x}^{3}}-7{{x}^{2}}-5x+11$, $x\in \mathbb{R}$, $x\ge k$,</p>
<p style="margin-left: 9rem">$\mathrm{g}:x\mapsto {{\left( x+1 \right)}^{2}}+2$,$x\in \mathbb{R}$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Let&nbsp;$k=1$.</div><ol class="abc-list"><li><div class="list-item-paragraph">Show that&nbsp;$\mathrm{f}$ does not have an inverse.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Determine whether the composite function $\mathrm{fg}$ exists.<span class="inline-marks">[2]</span></div></li></ol></li><li><div class="list-item-paragraph">Find the value of $k$ given that&nbsp;${{\mathrm{f}}^{-1}}$ exists and that the domain of ${{\mathrm{f}}^{-1}}$ is $x\in \mathbb{R}$, $x\ge -24$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Let&nbsp;$k=6$.</div><ol class="abc-list"><li><div class="list-item-paragraph">Show algebraically that&nbsp;$\mathrm{f}'(x)&gt;0$ for all values of $x$ in the domain of $\mathrm{f}$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Solve the equation $\mathrm{g}{{\mathrm{f}}^{-1}}(x)=83$.<span class="inline-marks">[3]</span></div></li></ol></li></ol>

Functions $\mathrm{f}$ and $\mathrm{g}$ are defined by

$\mathrm{f}:x\mapsto {{x}^{3}}-7{{x}^{2}}-5x+11$, $x\in \mathbb{R}$, $x\ge k$,

$\mathrm{g}:x\mapsto {{\left( x+1 \right)}^{2}}+2$,$x\in \mathbb{R}$.

Find the value of $k$ given that ${{\mathrm{f}}^{-1}}$ exists and that the domain of ${{\mathrm{f}}^{-1}}$ is $x\in \mathbb{R}$, $x\ge -24$.

Modulus Functions

Functions

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

2020 EJC Promo Q10

$f(x) = x^3 - 7x^2 - 5x + 11$, $x \in \mathbb{R}$, $x \geq 1$
The line $y = -10$ cuts $f$ twice, hence $f$ is not a one-one function.
Thus, $f^{-1}$ doesn't exist.

$g(x) = (x + 1)^2 + 2$, min $(-1, 2)$
For $fg$ to exist, $R_g \subseteq D_f$
$R_g = [2, \infty)$, $D_f = [1, \infty)$
Since $R_g \subseteq D_f$, $\therefore fg$ exists.

$f^{-1}$ exists and $D_{f^{-1}} = [-24, \infty) = R_f$
When $y = -24$, $x = 7$ or $-2.236$ (rej as $f^{-1}$ exists)
$\therefore k = 7$

$f(x) = x^3 - 7x^2 - 5x + 11$
Need to show $f'(x) > 0$ for $x \geq 6$
$f'(x) = 3x^2 - 14x - 5$
$= 3\left[x^2 - \frac{14}{3}x\right] - 5$
$= 3\left[\left(x - \frac{7}{3}\right)^2 - \left(\frac{7}{3}\right)^2\right] - 5$
$= 3\left(x - \frac{7}{3}\right)^2 - \frac{64}{3}$
When $k = 6$, $D_f = [6, \infty)$
$\therefore x \geq 6$
$\left(x - \frac{7}{3}\right)^2 \geq \left(6 - \frac{7}{3}\right)^2 = \frac{121}{9}$
$3\left(x - \frac{7}{3}\right)^2 \geq 3\left(\frac{121}{9}\right)$
$3\left(x - \frac{7}{3}\right)^2 - \frac{64}{3} \geq 3\left(\frac{121}{9}\right) - \frac{64}{3}$
$f'(x) \geq 19 > 0$
$\therefore f'(x) > 0$ (shown)

$g[f^{-1}(x)] = 83$
$f(x) = x^3 - 7x^2 - 5x + 11$, $x \geq 6$
$g(x) = (x + 1)^2 + 2$, $x \in \mathbb{R}$
Let $a = f^{-1}(x) \Rightarrow f(a) = x \Rightarrow a \geq 6$
$g(a) = 83$
$(a + 1)^2 + 2 = 83$
$(a + 1)^2 = 81$
$a + 1 = \pm 9$
$a = -1 + 9 = 8$ or $a = -1 - 9 = -10$ (rej as $a \geq 6$)
$a = 8$
$\therefore 8 = f^{-1}(x)$
$f(8) = x$
$x = 8^3 - 7(8)^2 - 5(8) + 11$
$x = 35$

2020 EJC Promo Q10

2020 EJC Promo Q10

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