Solving for Maximum Value of h in \mu = \lambda h

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Homework Statement



I'm asked to find the maximum value of h such that \mu = \lambda h\;\;(\lambda \in \mathbb{R}) satisfies: |1+\mu + \frac{1}{2} \mu ^2 + \frac{1}{6} \mu ^3| < 1

The Attempt at a Solution



My hurdle is solving 1+\mu + \frac{1}{2} \mu ^2 + \frac{1}{6} \mu ^3=1 to find the interval \mu\in (?,?) which satisfies |1+\mu + \frac{1}{2} \mu ^2 + \frac{1}{6} \mu ^3| < 1
 
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-1 from both sides, then obviously μ=0 is a solution, and from there you should be able to find the rest of the intervals using the fact it's a positive cubic.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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