Find a number such that the sum of the number and its square will be as small as possible. Find a number such that the sum of the number and its square will be as small as possible.
Please try to explain as well as you can.
 A: Let the number be $x$. Then you want to find $x$ such that $x+x^2$ is minimum.
Note that you can write $x+x^2$ as $\left( x+ \dfrac12\right)^2 - \dfrac14$.
Can you finish it from here, by noting that a square term is always non-negative?
Move your mouse over the gray area to see the complete answer.

 As stated above, a square term is always non-negative. Hence, we have that $\left( x+ \dfrac12\right)^2 \geq 0$. Adding $-\dfrac14$ to both sides, we get that $x + x^2 = \left( x+ \dfrac12\right)^2 - \dfrac14 \geq -\dfrac14$. Hence, the minimum value is $-\dfrac14$ and is attained when $\left( x+ \dfrac12\right)^2 = 0$ i.e. when $\left( x+ \dfrac12\right) = 0$. Hence the minimum value is $-\dfrac14$ and is attained at $x = -\dfrac12$.

A: Here is a solution using derivatives.
We want to find the number $x$ that minimizes $x^2 + x$, the derivative of which is $2x + 1$. To find critical points, we solve $2x+1 = 0$ to get $x = -1/2$.
So far, we have shown only that $x = 1/2$ is either a local maximum or a local minimum of $x^2 + x$. Considering the shape of the graph, however, we know that it is a parabola opening upward, and so has no local maxima. Thus, $x = -1/2$ is our desired minimum value.
(Another way to distinguish between local maxima and minima is to consider the second derivative, which is the constant function $2$ in this case. Since this is positive for all $x$, the graph of $x^2 + x$ is concave up for all $x$. Thus, we again conclude that $x = -1/2$ must be a local minimum.)
