Find the minimum and maximum values of f(x) =

Find the minimum and maximum values of $f(x) = \sqrt{2ax^{2}\bigl(2a-x\bigr)}$, where $a\gt 0$.

Step 1: Domain and sign

Since $a\gt 0$ and $x^{2}\ge0$, the radicand $2ax^{2}\bigl(2a-x\bigr)$ is non-negative exactly when $2a-x\ge0$. The domain is therefore $\bigl(-\infty,\,2a\bigr]$, and on it $f\ge0$, with $f(x) = 0$ only at $x = 0$ and $x = 2a$.

Step 2: Behaviour on the whole domain

For $x\to-\infty$ both factors $x^{2}$ and $2a-x$ grow, so $f(x)\sim\sqrt{2a}\,|x|^{3/2}\to+\infty$: the function is unbounded above. Hence on $\bigl(-\infty,2a\bigr]$ there is no greatest value, while the least value is $0$, attained at $x = 0$ and at $x = 2a$. The interesting question is the maximum on the finite interval $\bigl[0,2a\bigr]$ bounded by the two zeros, and that is what follows.

Step 3: The derivative for $0\lt x\lt 2a$

There $f(x) = \sqrt{2a}\;x\,\sqrt{2a-x}$, so by the product rule

\[ f'(x) = \sqrt{2a}\left(\sqrt{2a-x}+x\cdot\frac{-1}{2\sqrt{2a-x}}\right) = \sqrt{2a}\cdot\frac{2\bigl(2a-x\bigr)-x}{2\sqrt{2a-x}} = \sqrt{2a}\cdot\frac{4a-3x}{2\sqrt{2a-x}}. \tag{1} \]

Step 4: The critical point

The denominator in (1) is positive for $x\lt 2a$, so $f'(x) = 0$ exactly when

\[ 4a-3x = 0 \;\Longrightarrow\; x = \frac{4a}{3}, \]

and $\dfrac{4a}{3}\lt 2a$, so the point lies inside the interval.

Step 5: Sign analysis

By (1) the sign of $f'$ is the sign of $4a-3x$: \[0\lt x\lt \frac{4a}{3}:\ f'\gt 0\ \ (\text{increasing}); \qquad \frac{4a}{3}\lt x\lt 2a:\ f'\lt 0\ \ (\text{decreasing}).\]

The change from $+$ to $-$ marks a maximum.

Step 6: The maximum value

\[ f\!\left(\frac{4a}{3}\right) = \sqrt{2a\cdot\frac{16a^{2}}{9}\cdot\left(2a-\frac{4a}{3}\right)} = \sqrt{2a\cdot\frac{16a^{2}}{9}\cdot\frac{2a}{3}} = \sqrt{\frac{64a^{4}}{27}} = \frac{8a^{2}}{3\sqrt3} = \frac{8\sqrt3\,a^{2}}{9}. \tag{2} \]

Step 7: Verification

\[a = 1:\quad f\!\left(\tfrac43\right) = 1.5396007 = \frac{8\sqrt3}{9}; \qquad f(1) = 1.4142136; \qquad f(1.5) = 1.5 \;\;\checkmark\]

\[f(0) = f(2) = 0,\ \text{and } \tfrac43 \text{ lies between them, as a maximum must} \;\;\checkmark\]

\[\text{Squaring: } f^{2} = 2a\bigl(2ax^{2}-x^{3}\bigr),\ \bigl(f^{2}\bigr)' = 2a\,x\bigl(4a-3x\bigr) = 0 \ \text{at } x = \tfrac{4a}{3} \;\;\checkmark\]