SOLUTION 7: We are given the equation
$$ x^3+2= \sin{x} \ \ \ \ \longrightarrow \ \ \ \ x^3+2-\sin{x}=0 $$
Let function
$$ f(x)=x^3+2-\sin{x} \ \ \ \ and \ choose \ \ \ \ m=0 $$
This function is continuous for all values of $x$ since it is the DIFFERENCE of continuous functions; it is well-known that $ \sin{x}$ is continuous for all values of $x$ and $x^3+2$ is continuous for all values of $x$ since it is a polynomial.
To establish an appropriate interval consider the graph of this function. (Please note that the graph of the function is not necessary for a valid proof, but the graph will help us understand how to use the Intermediate Value Theorem. On many subsequent problems, we will solve the problem without using the "luxury" of a graph.)
Note that $$ f(0)= (0)^3+2-\sin{(0)} =2>0 \ \ \ \ and \ \ \ \ f(-\pi)= (-\pi)^3+2-\sin{(-\pi)}= -\pi^3+2-0=2-\pi^3 \approx -29<0 $$
so that $$ f(-\pi) \approx -29 < m <2=f(0) $$
i.e., $m=0$ is between $ f(-\pi) $ and $ f(0) $. Now choose the interval to be $ \ [-\pi, 0] $.
The assumptions of the Intermediate Value Theorem have now been met, so we can conclude that there is some number $c$ in the interval $[-\pi, 0]$ which satisfies
$$ f(c)=m $$
i.e.,
$$ c^3+2-\sin{c}=0 $$
and the equation is solvable.
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