Finding the Extrema of a Function (without differetiation) $$
(t^2-t+1)/(t^2+t+1)
$$
prove that the function is upper bounded by 3 and lower bounded by 1/3 without differentiation
 A: 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 $\min A = \dfrac{1}{3}.$
A: 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.
