$2^{\left( x^{3}-x\right) } < 1$
Let $2^{\left( x^{3}-x\right) }-1=f\left( x\right)$
To find the values for which $f(x)<0$ I let $f(x)=0$:
$2^{\left( x^{3}-x\right) }-1=0$
$2^{\left( x^{3}-x\right) }=1$
$\ln \left( 2^{\left( x^{3}-x\right) }\right) =\ln \left( 1\right) $
$\left( x^{3}-x\right) \ln \left( 2\right) =\ln \left( 1\right) $
$\left( x^{3}-x\right) =\dfrac {\ln \left( 1\right) }{\ln \left( 2\right) }$
$x^{3}-x=0$
$x\left( x^{2}-1\right) =0$
So $f(x)=0$ when $x=0$, $x=1$ and $x=-1$
After testing values in intervals between those solutions I guess I lucked out and got the answer to be $f(x)<0$ on $\left( -\infty ,-1\right) \cup \left( 0,1\right) $
Although I got the write answer this time I'm not very confident in my ability to do so for other questions, is there a way to do this very "analytically" so you're absolutely sure of the answer and that it holds for all x in the given interval? What tends to be general method? You find the values for which f(x) is zero and take values around that? Are there any tricks you can use? Thanks.
$2^{\left( x^{3}-x\right) } < 2^0 \Rightarrow x^{3}-x < 0$
And if the base of the exponent is $<1$ then the inequality gets reversed (you know why).