How to prove that there is no interval that maps to itself under a function

Question

I have the function $ g(x) = x^3 + 3x^2 - 3 $ and I need to show that there is no interval $ [a,b] $ such that $ g:[a,b] \mapsto [a,b] $. How do I go about this?

Thanks a lot

Answer 1

Counterexample (fixed point): $$g : [1,1] \mapsto [1,1] $$

Answer 2

I infact found the answer to be incorrect. Consider the interval $[-3, 1]$. Claim:

$g([-3, 1]) = [-3,1]$.

Proof: $g(-3) = -3$, and $g(1) = 1$. Then $g'(x) = 3x^2 + 6x = 3x(x + 2) = 0 \iff x = 0, -2$.

$g(0) = -3$, and $g(-2) = 1$. So $g_{max} = g(-2) = g(1) = 1$, and $g_{min} = g(0) = g(-3) = -3$. So $g([-3, 1]) \subseteq [-3, 1]$. Now we prove: $[-3, 1] \subseteq g([-3, 1])$. Let $-3 < r < 1$, and consider $g(x) = r \iff x^3 + 3x^2 - 3 - r = 0$. $g(-3) = - 3 - r < 0$, and $g(1) = 1 - r > 0$, so by the intermediate value theorem: there is an $x \in (-3, 1)$ such that: $g(x) = r$. So this means: $g([-3, 1]) = [-3, 1]$.