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<p>The diagram shows the graphs of ${y = e^x - 3x}$. The two roots of the equation ${{\mathrm{e}}^{x}}-3x=0$ are denoted by ${\alpha}$ and ${\beta}$, where&nbsp;${\alpha &lt; \beta}$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Find the values of ${\alpha}$ and ${\beta}$, each correct to $3$ decimal places.</div></li></ol>
<p>A sequence of real numbers ${{x}_{1}}$, ${{x}_{2}}$, ${{x}_{3}}$, … satisfies the recurrence relation</p>
<p style="text-align: center">${{x}_{n+1}}=\frac{1}{3}{{\mathrm{e}}^{{{x}_{n}}}}$ for&nbsp;$n\ge 1$.</p>
<ol class="i-list" start="2"><li><div class="list-item-paragraph">Prove algebraically that, if the sequence converges, then it converges to either ${\alpha}$ or ${\beta}$.</div></li><li><div class="list-item-paragraph">Use a calculator to determine the behaviour of the sequence for each of the cases ${x_1 = 0}$, ${x_1 = 1}$, ${x_1 = 2}$.</div></li><li><div class="list-item-paragraph">By considering $x_{n+1}&lt;x_{n}$ prove that</div><ol class="abc-list"><li><div class="list-item-paragraph">$x_{n+1}&lt;x_n$ if&nbsp;$\alpha &lt; x_{n} &lt; \beta$,</div></li><li><div class="list-item-paragraph">$x_{n+1}&gt;x_n$ if $ x_n &lt; \alpha $ or $x_n&gt;\beta$.</div></li></ol></li><li><div class="list-item-paragraph">State briefly how the results in part <strong>(iv)</strong> relate to the behaviors determined in part <strong>(iii)</strong>. </div></li></ol>

The diagram shows the graphs of ${y = e^x - 3x}$. The two roots of the equation ${{\mathrm{e}}^{x}}-3x=0$ are denoted by ${\alpha}$ and ${\beta}$, where ${\alpha < \beta}$.

Find the values of ${\alpha}$ and ${\beta}$, each correct to $3$ decimal places.

A sequence of real numbers ${{x}_{1}}$, ${{x}_{2}}$, ${{x}_{3}}$, … satisfies the recurrence relation

${{x}_{n+1}}=\frac{1}{3}{{\mathrm{e}}^{{{x}_{n}}}}$ for $n\ge 1$.

Prove algebraically that, if the sequence converges, then it converges to either ${\alpha}$ or ${\beta}$.

Use a calculator to determine the behaviour of the sequence for each of the cases ${x_1 = 0}$, ${x_1 = 1}$, ${x_1 = 2}$.

By considering $x_{n+1}<x_{n}$ prove that

State briefly how the results in part **(iv)** relate to the behaviors determined in part **(iii)**. 

N2007 P1 Q9

N2007 P1 Q9

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