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

Question

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.

Answer 1

Hint. I'm giving you the derivative and let the rest for you. With the chain rule, one has

$$\left ( \frac{3}{x^2 + x + 1} \right)' = \big ( 3 (x^2 + x + 1)^{-1} \big )' = - 3 (x^2 + x + 1)^{-2} \cdot (2x + 1) = - \frac{3 (2x + 1)}{(x^2 + x + 1)^2}.$$

Answer 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.