Proving no real solutions Please read the question before marking as duplicate; this is proving there are NO SOLUTIONS OUTSIDE THE INTERVAL.


  Given that:
  
  $x^3−3x+3=\sin(x)$ show that there are no real solutions outside the
  interval $[−3,−2]$


am not trying to prove there is a solution; this has been done with intermediate value theorem; proving no real solutions outside interval
By derivation, the LHS has extrema at $x=\pm1$, namely $(-1,5)$ and $(1,1)$. We also have the points $(-3,-15)$ and $(-2,1)$.

We can write the following variation table:

$$\begin{array}&amp;x&amp;-\infty&amp;-3&amp;-2&amp;-1&amp;1&amp;\infty\\LHS&amp;-\infty&amp;-15&amp;1&amp;5&amp;1&amp;\infty\end{array}$$

This proves that the LHS remains out of $[-1,1]$ outside $[-3,-2]$, except maybe at $x=1$. But then $\sin 1&lt;1$ and there is no solution.
Nice! How's question 3c doing for you?

Essentially what you want to do is find the critical points and prove that the local minimum does not occur below the x-axis.
You would find the critical points by deriving f(x) to get f^(1)(x) = 3x^2-3-cos(x).
You know that -1&lt;=cos(x)&lt;=1, so in order for f^(-1)(x) to equal 0, the component |3x^2-3|&lt;=1. Solving this, x&lt;= sqrt(4/3), and x>=-sqrt(4/3)
Find the derivative of both, use either first or second derivative principals to determine nature, and check endpoints to see if they are strictly increasing, or strictly decreasing.

Simple...ish