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$$ (t^2-t+1)/(t^2+t+1) $$

prove that the function is upper bounded by 3 and lower bounded by 1/3 without differentiation

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2 Answers 2

Let the fraction be $f(t)$. Look at the differences $3-f(t)$ and $f(t)-\frac13$; showing that they are non-negative for all $t$ shows that $\frac13\le f(t)\le 3$ for all $t$, and showing that they can be $0$ shows that these bounds are the best possible. For instance,

$$3-\frac{t^2-t+1}{t^2+t+1}=\frac{2t^2+4t+2}{t^2+t+1}=\frac{2(t+1)^2}{\left(t+\frac12\right)^2+\frac34}\;,$$

which is clearly never negative and is $0$ when $t=-1$. Thus, $f(t)$ is always less than or equal to $3$, and it’s equal when $t=1$. Now try the same sort of calculation for $1/3$.

This doesn’t show how to find the bounds without calculus, but it does show how to verify them.

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thanks so much! –  Paul Smith Sep 28 '12 at 8:11
    
Ya, i got that, the question asked us to "verify". That's a simple way. –  Paul Smith Sep 28 '12 at 8:23

Putting $A=\dfrac{t^2-t+1}{t^2+t+1}.$ And take the equation has the form $$(A-1)t^2+(1+A)t+A-1 = 0.$$ 1) With $A = 1$, the equation is always has solution. 2) With $A \neq 1$, the equation has solution when and only when $$-3A^2+10A-3\geqslant 0$$ and then we get $\dfrac{1}{3} \leqslant A \leqslant 3.$ Thus, $\max A = 3$ and $\min A = \dfrac{1}{3}.$

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thanks so much! –  Paul Smith Sep 28 '12 at 8:09
    
But why is it that there must be a solution? –  Paul Smith Sep 28 '12 at 8:21

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