2015/006) Find the maximum value of $P(x)=x^3 −3x$ on the set of all real x that satisfies the inequality $x^4 −13x^2 +36\le0$.

we have $x^4-13x^2+ 36 \le 0$
or $(x^2-9)(x^2-4) \le 0$
or $4\le x^2\le 9$

so we get 2 ranges for x

1) $-3\le x\le -2$
2)  $2\le x\le 3$

we need to find the maximum of $x^3-3x$ in this ranges

$x^3-3x = x(x^2-3)$

in the 1st range x is negative  and x^2-3 is positive so in the range $x^3-3x$ is -ve

for $x\gt \sqrt{3}$ and hence for $x \ge 2$ both x and x^2- 3 are positive and increasing so the highest value is at highest x that is x = 3 so the value of $x^3-3x=18$