Maxima and minima of $\frac{x^2-3x+4}{x^2+3x+4}$ without calculus

Question

How to find the minimum and max values of $y=\frac{x^2-3x+4}{x^2+3x+4}$ for all real values of $x$ without using calculus?

Perhaps it could be done graphically by noting the fact that the numerator and denominator are a pair of parabolas symmetric about the $x$ axis, but I do not know how to continue.

Thanks!

Answer 1

Let the maximum of $f(x) = \frac{x^2-3x+4}{x^2+3x+4}$ be $m$. Then:

$$x^2-3x+4 = mx^2 + 3mx + 4m$$ $$(m-1)x^2 + (3m+3)x + (4m - 4) = 0$$

We want this equation to have only one real root (a double root), so: $$\Delta = 0 \Rightarrow (3m+3)^2-4(m-1)(4m-4) = 0.$$

A similar process for the minimum ($n$) yields the same equation, as multiplying it by $-1$ does not change the values of $m$. Therefore, both the maximum and minimum values are given by this equation.

Answer 2

Hint wehave $$\frac{x^2-3x+4}{x^2+3x+4}=y$$ $$x^2(y-1)+x(3y+3)+4y-4=0$$ set the discriminant greater or equal to zero

Answer 3

$$\dfrac{x^2-3x+4}{x^2+3x+4}=1-\dfrac{6x}{x^2+3x+4}=1-\dfrac6{x+\dfrac4x+3}$$

Now if $x>0, x+\dfrac4x\ge2\sqrt{x\cdot\dfrac4x}=4$

$\implies\dfrac1{x+\dfrac4x+3}\le\dfrac17\implies-\dfrac6{x+\dfrac4x+3}\ge-\dfrac67$

If $x<0, x=-y, y>0, x+\dfrac4x=-\left(y+\dfrac4y\right)$

Can you take it from here?

Answer 4

They did not specify that calculus mustn't be used, but I was curious if it could be solved in a simpler way – Monocerotis Nov 20 at 8:19

thanks man, you saved me from a lot of differentiation and substitution – Monocerotis Nov 20 at 8:46

The most simple way to solve your problem is by using calculus:

By applying the quotient rule, you get: $y'(x)=(\frac{x^2-3x+4}{x^2+3x+4})'=\frac{(2x-3)(x^2+3x+4)-(2x+3)(x^2-3x+4)}{(x^2+3x+4)^2}$.

After putting the condition $y'(x)=0$ and expanding the numerator of $y'(x)$, you obtain:

$x^2-4=0$, whose solutions are:

$x_1=2$ and $x_2=-2$.

In conclusion:

$y_{max}=7$ (for $x=-2$) and $y_{min}=\frac{1}{7}$ (for $x=2$).