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The maximum value of $f(x)=x^{n}(1 - x)^{n}$ for a natural numbers $n\geq 1$ and $0\leq x\leq 1$ is

1. $\frac{1}{2^{n}}$
2. $\frac{1}{3^{n}}$
3. $\frac{1}{5^{n}}$
4. $\frac{1}{4^{n}}$
asked in Calculus | 210 views

\begin{align}f(x) &= x^n (1-x)^n \\[2em] \frac{d}{dx}\,f(x) &= nx^{n-1}(1-x)^n - nx^n(1-x)^{n-1} \\[1em] &= nx^{n-1}(1-x)^{n-1} \Bigl ( (1-x) - x \Bigr ) \\[1em] &= n x^{n-1} (1-x)^{n-1} (1-2x) \end{align}

Putting $\frac{d}{dx}\,f(x) = 0$ to get the critical points, we get:

\left \{ \begin{align}x &= 0 \quad \text{or},\\(1-x) &= 0 \quad \text{or},\\ (1-2x) &= 0 \quad \text{or},\\n&=0 \end{align} \right.

$x = 0$ and $(1-x) = 0$ are the endpoints, which give $f(x) = 0$, hence the minima.

$n = 0$ can't be assumed since we are given that $n>0$, and $n$ is a parameter, not a decision variable.

$(1-2x) = 0$ gives us our maximum.

Thus, the maximum point is at $x = \frac1 2$ which makes $f(x) = \frac1{4^n}$.

Option d) is the correct answer.

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