Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Hint: You are trying to find the minimum of $x^2+x$. Try expressing this as a completed square: $(x+1/2)^2-1/4$ and see what you can conclude. –  Old John Jun 10 '12 at 21:55
1  
Why have you tagged this trigonometry? I have removed the trigonometry tag and retagged it as algebra-precalculus. –  user17762 Jun 10 '12 at 21:56
    
Apologies, @Marvis. I suggested in a previous post that the OP also tag what class the homework was for, to give us so.e context. –  Cameron Buie Jun 10 '12 at 22:58

2 Answers 2

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 $x = -\dfrac12$.

share|improve this answer

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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.