What Is the Maximum Value of P(x) on the Interval Where x⁴-13x²+36≤0?

[anemone]

Find, with explanation the maximum value of $P(x)=x^3-3x$ on the set of all real $x$ that satisfies the inequality $x^4-13x^2+36\le 0$.

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[anemone]

Congratulations to the following members for their correct solutions::)

1. kaliprasad
2. greg1313

Solution from greg1313:

Spoiler

$$x^4-13x^2+36=0\implies x^2=\frac{13\pm5}{2}\implies x=\pm2,\pm3$$so the inequality$$x^4-13x^2+36\le0$$is satisfied when $$x\in[-3,-2]\cup[2,3]$$If$$P(x)=x^3-3x$$then$$P'(x)=3x^2-3$$which has zeros at $x=\pm1$$$P''(x)=6x$$so, by the second derivative test, $P(x)$ has a local maximum at $-1$ and a local minimum at $1$.$$\text{ }$$Using the properties of the cubic function we can determine that$$\text{ }$$$P(-3)\lt P(-2)\lt P(-1)=2$ and $P(2)\lt P(3)=18$ so, over the set $[-3,-2]\cup[2,3]$, $\max\left(P(x)\right)=18$.