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On the interval $[0,1]$ the function $x^{25}(1-x)^{75}$ takes its maximum value at the point
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Verified Answer
The correct answer is:
$1 / 4$
Given, $f(x)=x^{25}(1-x)^{75}$
$$
\begin{aligned}
\Rightarrow f^{\prime}(x) &=25 x^{24}(1-x)^{75}-75 x^{25}(1-x)^{74} \\
&=25 x^{24}(1-x)^{74}(1-4 x) \\
\therefore \quad f^{\prime}(x) &=0 \\
\Rightarrow \quad x &=0,1,1 / 4
\end{aligned}
$$
If $x < 1 / 4$, then
$$
f^{\prime}(x)=25 x^{24}(1-x)^{74}(1-4 x)>0 .
$$
and if $x>1 / 4$, then
$$
f^{\prime}(x)=25 x^{24}(1-x)^{74}(1-4 x) < 0 .
$$
Thus, $f^{\prime}(x)$ changes its sign from positive to negative as $x$ passes through $1 / 4$ from left to right. Hence, $f(x)$ attains its maximum at $x=1 / 4$.
$$
\begin{aligned}
\Rightarrow f^{\prime}(x) &=25 x^{24}(1-x)^{75}-75 x^{25}(1-x)^{74} \\
&=25 x^{24}(1-x)^{74}(1-4 x) \\
\therefore \quad f^{\prime}(x) &=0 \\
\Rightarrow \quad x &=0,1,1 / 4
\end{aligned}
$$
If $x < 1 / 4$, then
$$
f^{\prime}(x)=25 x^{24}(1-x)^{74}(1-4 x)>0 .
$$
and if $x>1 / 4$, then
$$
f^{\prime}(x)=25 x^{24}(1-x)^{74}(1-4 x) < 0 .
$$
Thus, $f^{\prime}(x)$ changes its sign from positive to negative as $x$ passes through $1 / 4$ from left to right. Hence, $f(x)$ attains its maximum at $x=1 / 4$.
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