# Find the maxima and minima of $y = \frac{3}{x^2 + x + 1}$

This is from "Calculus Made Easy", Exercises 10, problem 5, on page 137.

I've taken the derivative and reduced it to:

$$2x^3 + 3x^2 + 3x + 1 = 0$$

But I'm lost on finding the roots of a cubic.

• Your derivative is wrong. – Martin R Aug 3 '19 at 17:14
• Your derivative is incorrect. Write the function in the form $y = 3(x^2 + x + 1)^{-1}$, then take the derivative. – N. F. Taussig Aug 3 '19 at 17:14
• I suggest you complete the square for $p(x)=x^2+x+1$ and look at that expression to see what can be done. When is $p(x)$ minimal and what does it do for $1/p(x)$? – mf67 Aug 3 '19 at 17:15
• Woop, I think I see my mistake. – Mike Aug 3 '19 at 17:28
• Yep, when I correctly differentiate I get $x = -\frac{1}{2}$, which is correct. – Mike Aug 3 '19 at 17:32

$$\left ( \frac{3}{x^2 + x + 1} \right)' = \big ( 3 (x^2 + x + 1)^{-1} \big )' = - 3 (x^2 + x + 1)^{-2} \cdot (2x + 1) = - \frac{3 (2x + 1)}{(x^2 + x + 1)^2}.$$
It is not good to use derivatives everywhere ! Here in this problem we can use the fact that $$x^{2}+x+1 = (x+ \frac {1}{2})^{2} + \frac {3}{4}$$ and we can see that it is positive for all $$x$$ belonging to the real number system. Hence maxima will be attained when the denominator is the minimum which can be attained only when $$x = \frac {-1}{2}$$ you can work for the minima to in a similiar manner it would turn out that you can find infimum=0 but there is no exact minimum.
• I didn't get you 😂? What is there to understand I used $\frac {1}{4} + \frac {3}{4} =1$ and assumed you know expansion of $(a+b)²$ – Aditya Garg Aug 3 '19 at 21:29