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How many answers does this equation have ? (using graphing functions)

$$x^3+3=\sin{x}-x$$ Book solution:We draw both $x^3+x+3$ and $\sin{x}$ and we see that they intersect only once?

Now I don't know why they intersect only once?

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    What is gragh/graghing? – 2017-01-09
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    I suppose he means "graph" – 2017-01-09

3 Answers 3

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

$y= \sin x$

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    Ok but you have drawn it using a program not by hand. – 2017-01-09
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    @TahaAkbari: If graphing by program or by hand is an important consideration, you should have said so in your question. – 2017-01-09
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Without "graghing":

Let $f(x)=x^3+x+3- \sin x$. Then $f(0)=3>0$ and $f( - \pi)= - \pi^3- \pi +3<0$.

hence there is $x_0$ such that $f(x_0)=0$.

Since $f'(x)=3x^2+1- \cos x >0$ for all $x \ne 0$, $f'$ is strictly increasing, thus there is exactly one $x_0$ such that $f(x_0)=0$.

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The functions are continuous and differentiable. Taking the derivative of the difference of the members,

$$3x^2-\cos x+1$$ is a positive expression, hence there can be only one root. This root indeed exists because

$$-3\cdot\infty^3-\infty+3<-\sin\infty\text{ and }3\cdot0^2+0+3>\sin 0.$$

(If you don't like to evaluate at $-\infty$, use $-\pi$).