Friday, October 28th

Problem
Find, with explanation, the maximum value of $f(x)=x^3-3x$ on the set of all real numbers $x$ satisfying $x^4 +36 \leq 13x^2$.

Solution

As the function is continuous, we only need to check for local maxima and boundary points to find the maximum. Solving the inequality gives

$x^4-13x^2+36\leq 0$

hence $x^2 \in [4,9]$ so $x\in [-3,-2]\cup [2,3]$. As the only local maximum of $f(x)$ is at $x=-1$, we only need to check for boundaries. Then the max is at $x=3$.