How many answers does the equation $x^3+3=\sin{x}-x$ have (using graphing)?

Question

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?

Answer 1

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

$y= \sin x$

Answer 2

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 $f(x_0)=0$.

Answer 3

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 $-\pi$).