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How doese one solve this equation?

$$ 3x+\sin x = e^x $$

I tried graphing it and could only find approximate solutions, not the exact solutions. My friends said to use Newton-Raphson, Lagrange interpolation, etc., but I don't know what these are as they are much beyond the high school syllabus.

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Problems like these generally only have numerical answers. Graph it. – Qiaochu Yuan Jun 28 '12 at 13:03
@Bazinga: graph $y = e^x - 3x - \sin x$ with a calculator (or WolframAlpha I suppose). The points at which it crosses the $x$-axis are your solutions. – Qiaochu Yuan Jun 28 '12 at 13:13
Bazinga, do you like programming? – Yrogirg Jun 28 '12 at 13:14
Wow, you must have a rough teacher if this was in your exam. – Neal Jun 28 '12 at 13:31
@Bazinga: you will never get asked this question on an exam. – Qiaochu Yuan Jun 28 '12 at 13:37
up vote 5 down vote accepted

A nice method to find an approximate solution is to successively cut intervals in half, as follows: let's first rewrite this as $$f(x) = 3x + \sin x - e^x = 0$$ Now pick two values, $a$ and $b$, such that $f(a) < 0$ and $f(b) > 0$. (You might have to make a few guesses before finding such values!) In this case, let's choose $a = 0$ and $b = 1$: $$f(a) = 3(0)+\sin(0)-e^{0} = -1 < 0$$ $$f(b) = 3(1)+\sin(1)-e^{1} = 1.12... > 0$$

Now, because our function $f(x)$ is "smooth", there must be a solution somewhere between $a$ and $b$. Find the point halfway in between them, $\frac{0+1}{2} = 0.5$, and check to see whether it makes $f(x)$ positive or negative: $$f(0.5) =3(0.5)+\sin(0.5)-e^{0.5} = 0.33... > 0$$ So, since $f(x)$ is positive here, set $0.5$ as the new value for $b$. Once again calculate the midpoint of $a$ and $b$ (in this case, $\frac{0+0.5}{2} = 0.25$), evaluate it in $f$, and so on. Continue until you have the precision you are looking for.

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Here is the plot made by maple: – Sigur Jun 28 '12 at 23:30
how does this differ from fixed-point method versus your method of estimation? – User69127 Mar 17 '15 at 19:33
@User69127 What do you mean by "fixed-point method"? – Théophile Mar 17 '15 at 19:46
@User69127 I see. This method, the bisection method, is a naïve but reliable approach. It doesn't necessarily converge quickly, but it is guaranteed to converge. – Théophile Mar 18 '15 at 2:24


f(b)=3(1)+sin(1)−e1= is Not Equal to ---> 1.12...

f(b)=3(1)+sin(1)−e1= is Equal to ----> 0.299170578


f(0.5)=3(0.5)+sin(0.5)−e0.5= is Not Equal to ---> 0.33...>0

f(0.5)=3(0.5)+sin(0.5)−e0.5= is Equal to ----> -0.1399947352

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Welcome to MSE! It helps to format questions using MathJax (see FAQ). Also, it helps to provide details as to how you arrived at answers. Regards – Amzoti May 23 '13 at 13:11

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