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Let $x$ be a number which exceeds its square by the greatest possible quantity, then $x=$
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Verified Answer
The correct answer is:
$1 / 2$
Let the required numbers be $x$. Then,
$f(x)=x-x^{2}$
$\Rightarrow \quad f^{\prime}(x)=1-2 x$
For a local maxima or local minima, we must have
$$
\begin{aligned}
&\quad f^{\prime}(x)=0 \\
&\Rightarrow \quad 1-2 x=0 \\
&\Rightarrow \quad 2 x=1 \Rightarrow x=\frac{1}{2}
\end{aligned}
$$
Now, $\quad f^{\prime \prime}(x)=-2 < 0$
So, $x=\frac{1}{2}$ is point of local maxima.
Hence, required numbers is $\frac{1}{2}$.
$f(x)=x-x^{2}$
$\Rightarrow \quad f^{\prime}(x)=1-2 x$
For a local maxima or local minima, we must have
$$
\begin{aligned}
&\quad f^{\prime}(x)=0 \\
&\Rightarrow \quad 1-2 x=0 \\
&\Rightarrow \quad 2 x=1 \Rightarrow x=\frac{1}{2}
\end{aligned}
$$
Now, $\quad f^{\prime \prime}(x)=-2 < 0$
So, $x=\frac{1}{2}$ is point of local maxima.
Hence, required numbers is $\frac{1}{2}$.
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