Function problem!

$f'(x) = \frac{1}{x} + x - 2k$ . First solve $\frac{1}{x} + x - 2k = 0 \implies 1 + x^2 - 2kx = 0 \implies (x-k)^2 + 1 - k^2 = 0 \implies x = k \pm \sqrt{k^2 - 1}$ . For the function to have one local minimum and one maximum, there must be two real values of $x$ that satisfy the previous equation, which implies that $k^2 - 1 > 0 \implies k > 1$ . For now on, I will only consider $k \in (1,\infty)$ . Under these conditions, it is easy to check that $f''( k + \sqrt{k^2 - 1}) > 0$ and $f''(k - \sqrt{k^2 - 1}) < 0$ , showing that $k - \sqrt{k^2 - 1}$ is the local maximum.

Define $g(k) = k - \sqrt{k^2 - 1}$ . It is important to remark the property $f'(g(k)) = 0$ . The function $f(g(k))$ computes the maximum value of the function for a given $k$ , and we want to prove that $f(g(k)) < -2$ . For this, we can rely on calculus again. Compute $[f(g(k))]' = g'(k) f'(g(k)) - 2g(k) = -2g(k)$ .

For $k = 1, -2g(k) = -2$ , and as $\sqrt{k^2 - 1} < \sqrt{k^2} = k$ , $g(k) > 0 \implies -2g(k) < 0$ which implies that $f(g(k))$ is decreasing below its value at $k=1$ which is $-2$ .