Proving no real solutions

Question

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

Answer 1

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:

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

This proves that the LHS remains out of $[-1,1]$ outside $[-3,-2]$, except maybe at $x=1$. But then $\sin 1<1$ and there is no solution.

Answer 2

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<=cos(x)<=1, so in order for f^(-1)(x) to equal 0, the component |3x^2-3|<=1. Solving this, x<= 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