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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.
by Platinum (108k points)

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