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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.

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    $\begingroup$ Your derivative is wrong. $\endgroup$ – Martin R Aug 3 '19 at 17:14
  • $\begingroup$ Your derivative is incorrect. Write the function in the form $y = 3(x^2 + x + 1)^{-1}$, then take the derivative. $\endgroup$ – N. F. Taussig Aug 3 '19 at 17:14
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    $\begingroup$ 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 $1/p(x)$? $\endgroup$ – mf67 Aug 3 '19 at 17:15
  • $\begingroup$ Woop, I think I see my mistake. $\endgroup$ – Mike Aug 3 '19 at 17:28
  • $\begingroup$ Yep, when I correctly differentiate I get $x = -\frac{1}{2}$, which is correct. $\endgroup$ – Mike Aug 3 '19 at 17:32
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Hint. I'm giving you the derivative and let the rest for you. With the chain rule, one has

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

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  • $\begingroup$ Yeah, I messed up on the derivative by not reducing the -1 exponent to -2. $\endgroup$ – Mike Aug 3 '19 at 17:29
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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.

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  • $\begingroup$ This answer confuses me, as I don't see how you achieved the right half of the equation. $\endgroup$ – Mike Aug 3 '19 at 17:36
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    $\begingroup$ 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 $ $\endgroup$ – Aditya Garg Aug 3 '19 at 21:29
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    $\begingroup$ Okay, I see it now. Thanks. $\endgroup$ – Mike Aug 5 '19 at 1:48
  • $\begingroup$ Your welcome bro! $\endgroup$ – Aditya Garg Aug 5 '19 at 2:18

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