a. A critical point at x = 25 would mean f'(25) = 0 or doesn't exist. We have derivative
[tex]f(x) = x - k \sqrt x \implies f'(x) = 1 - \dfrac k{2\sqrt x}[/tex]
so that
[tex]f'(25) = 1 - \dfrac k{2\sqrt{25}} = 1 - \dfrac k{10} = 0 \implies \boxed{k=10}[/tex]
b. Taking the second derivative, we get
[tex]f''(x) = \dfrac k{4 x^{3/2}} = \dfrac5{2 x^{3/2}}[/tex]
At x = 25, the second derivative has a positive sign,
[tex]f''(25) = \dfrac 5{2 \times 25^{3/2}} = \dfrac 5{2\times5^3} = \dfrac1{50} > 0[/tex]
which means f(x) is concave upward around x = 25, so this critical point is a local minimum.
