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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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    Well we know $x\neq 1$ because e<3. we also know $x\neq 2$ because e^2 > 7 but \sin 2 < 1. And from there, we know that x\not> 2 because $e^x$ grows much more quickly as opposed to $3x + \sin x$. Therefore, we know that $x\in(1, 2)$. This may serve as a hint :)2018-02-27

2 Answers 2

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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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    @User69127 I see. This method, the [bisection method](http://en.wikipedia.org/wiki/Bisection_method), is a naïve but reliable approach. It doesn't necessarily converge quickly, but it is guaranteed to converge.2015-03-18
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sorry

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

And

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. Regards2013-05-23