How to find local maxima of $\frac{x^2+2x-3}{x^2+1}$ quickly?

Question

What is the value of relative maximum of the function $f(x)=\dfrac{x^2+2x-3}{x^2+1}$ ?

$1)-1+\sqrt5\qquad\qquad2)1+\sqrt5\qquad\qquad3)-1+\sqrt3\qquad\qquad4)1+\sqrt3$

It is a problem from a timed exam so I'm looking for the fastest approaches. Here is my approach:

First I tried taking derivative of the function and equate it to zero, but I realized by doing that I should solve a third degree equation because the degree of numerator is $2$. So I decided to write the fraction as:

$$f(x)=\dfrac{x^2+1+2x-4}{x^2+1}=1+\frac{2x-4}{x^2+1}$$ Hence it is easier to take derivative:

$$f'(x)=0\Rightarrow f'(x)=\dfrac{2x^2+2-4x^2+8x}{x^2+1}=0$$So we have $-x^2+4x+1=0\Rightarrow x=2\pm\sqrt5$. And from here by using Wolfram Alpha I got maximum is $\sqrt5-1$ at $2+\sqrt5$ and minimum is $-1-\sqrt5$ at $2-\sqrt5$.

But this approach takes much time (specially at the end) and it is not suitable for these kinds of exams. So can you please solve this problem differently?

Answer 1

$$\frac{2x-4}{x^2+1}=u$$

$$\begin{align} &\implies ux^2-2x+(u+4)=0\\ &\implies \Delta=1-u(u+4)≥0 \\ &\implies u^2+4u-1≤0 \\ &\implies (u+2)^2-5≤0 \\ &\implies |u+2|≤\sqrt 5 \\ &\implies \max \left\{u\right\}=\sqrt 5-2 \\ &\implies \max \left\{f(x)\right\}=\sqrt 5-1.\end{align}$$

Answer 2

I agree with others that you've probably computed the derivative as efficiently as possible. Perhaps an alternative approach may be not to try to do calculus faster, but to consider the properties of the function. The horizontal asymptote of the function is $ \ y = 1 \ , $ as the result of your polynomial division shows. The "remainder term", $ \ \frac{2·(x-2)}{x^2+1} \ $ , is always defined since the denominator is always positive. We see from the factorization that this term is negative for $ \ x < 2 \ $ and positive for $ \ x > 2 \ , $ so the curve of our function lies "below" the asymptote for $ \ x < 2 \ . $ We thus want to look for the relative maximum in the interval $ \ x > 2 \ . $ (This also eliminates choice $ (3) \ . ) $

Because a positive number is subtracted in the numerator and added in the denominator, we can write the inequality $ \ \frac{2x-4}{x^2+1} \ < \ \frac{2}{x} \ $ for $ \ x > 2 \ $ and hence

$$ \frac{x^2 + 2x - 3}{x^2 + 1 } \ < \ 1 + \frac{2}{x} \ \ \text{for} \ x > 2 \ \ . $$

The local (and in fact absolute) maximum of the function is therefore less than $ \ 1 + \frac22 \ = \ 2 \ , $ thereby eliminating choices $ \ (2) \ \ \text{and} \ \ (4) \ \ . $ This leaves $ \ \sqrt5 \ - \ 1 \ $ as the maximum value for the function. (A more leisurely calculation confirms this.)

I'm not sure working this "logic puzzle" is all that much faster than just using the Quotient Rule on the problem, but it is an alternative without having access to a calculator or graphing utility.