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Find all solutions of $e^{2 x}=x+6$, correct to 4 decimal places; use the Newton Method.
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Let $f(x)=e^{2 x}-x-6$. We want to find where $f(x)=0$. Note that $f^{\prime}(x)=2 e^{2 x}-1$, so the Newton Method iteration is
$x_{n+1}=x_{n}-\frac{e^{2 x_{n}}-x_{n}-6}{2 e^{2 x_{n}}-1}=\frac{\left(2 x_{n}-1\right) e^{2 x_{n}}+6}{2 e^{2 x_{n}}-1} .$
We need to choose an initial estimate $x_{0}$. This can be done in various ways. We can (if we are rich) use a graphing calculator or a graphing program to graph $y=f(x)$ and eyeball where the graph crosses the $x$-axis. Or else, if (like the writer) we are poor, we can play around with a cheap calculator, a slide rule, an abacus, or scrap paper and a dull pencil.

It is easy to verify that $f(1)$ is about $0.389$, and that $f(0.95)$ is about $-0.2641$, so by the Intermediate Value Theorem there is a root between $0.95$ and 1 . And since $f(0.95)$ is closer to 0 than is $f(1)$, maybe the root is closer to $0.95$ than to 1 . Let's make the initial estimate $x_{0}=0.97$.
The calculator then gives $x_{1}=0.970870836$, and then $x_{2}=0.97087002$. Since these two agree to 5 decimal places, we can perhaps conclude with some (but not complete) assurance that the root, to 4 decimal places, is $0.9709$. If we want greater assurance, we can compute $f(0.97085)$ and $f(0.97095)$ and hope for a sign change, which shows that there is a root between $0.97085$ and $0.97095$. There is indeed such a sign change: $f(0.97085)$ is about $-2.6 \times 10^{-4}$ while $f(0.97095)$ is about $10^{-3}$.

But the problem asked for all the solutions. Are there any others?

Draw the graphs of $y=e^{2 x}$ and $y=x+6$. The solutions of our equation are the $x$-coordinates of all places where the two curves meet. Even a rough picture makes it clear that the curves meet at some negative $x$. Since $e^{2 x}$ decays quite rapidly as $x$ decreases through negative values, it seems reasonable that there will be a single negative root, barely larger than $-6$. Certainly it cannot be smaller, since to the left of $-6, x+6$ is negative but $e^{2 x}$ is not. And it seems plausible that the positive root we found is the only one.

We first estimate the negative root. It is reasonable to start with $x_{0}=-6 .$ Then $x_{1}=-5.999993856 .$ We can guess that the root is indeed $-6$ to 4 decimal places. For certainty, we should check that $f(x)$ has different signs at $-6$ and $-5.9999$. It does.
Let's check that there are no more roots. Note that $f^{\prime}(x)=2 e^{2 x}-1$. Thus $f^{\prime}(x)>1$ when $x>0$, and in particular $f$ is increasing from $0 \mathrm{on}$, indeed it starts increasing at $x=-(1 / 2) \ln (2)$. Note also that $f(0)<0$, and that, for example, $f(1)>0$. So there is at least one root $r$ between 0 and 1 . But there can only be one root there. For $f(x)$ is increasing in the first quadrant, so can cross the $x$-axis only once.
A similar argument shows that there is a single negative root. For since $f(x)$ is negative in the interval $(-\infty,(1 / 2) \ln 2)$, the function $f$ is decreasing in this interval, so can cross the $x$-axis at most once in this interval. We saw already that it crosses the $x$-axis near $x=-6$.
Note. There are many other ways of solving the problem. For example our equation is equivalent to $2 x=\ln (x+6)$, and we could apply the Newton Method to $2 x-\ln (x+6)$. Or we can use basically the same approach as above, but let $y=2 x$. We end up solving $e^{y}=y / 2+6$. If we are doing the calculations by hand, this saves some arithmetic.
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