Animated Solution for Mathematics - Functions: If $f$ is an even function defined on the interval $(-5,5)$, then four real values of $x$ satisfying the equation $f(x)=f\left(\frac{x+1}{x+2}\right)$ are ........., ........., ........., and .........

Unlocking Even Functions

Imagine you are standing before a perfectly polished mirror. If you raise your right hand, your reflection raises its left. This is the essence of symmetry.

In the world of mathematics, an even function is exactly like that mirror. It is defined by the beautiful property that $f(x)=f(-x)$ for all $x$ in its domain.

Today, we are going to use this geometric intuition to solve a problem that might look intimidating at first glance but is actually a masterclass in logical deduction. We are given an even function $f$ defined on the interval $(-5,5)$, and we need to find the values of $x$ that satisfy the equation:

When we see an equation like $f(A)=f(B)$, our first instinct might be to simply equate $A$ and $B$. But wait! Because our function is even, there is a second, hidden path.

If $f(A)=f(B)$, then either $A=B$ (the inputs are identical) or $A=-B$ (the inputs are reflections of each other across the y-axis). This bifurcation is the heart of the problem.

We are not just solving one equation; we are exploring two distinct possibilities. Let us call the first input $A=x$ and the second input $B=\frac{x+1}{x+2}$.

The Identical Path

Let us start with the straightforward case: $A=B$. We set:

To solve this, we cross-multiply, which gives us $x(x+2)=x+1$. Expanding the left side, we get $x^2+2x=x+1$.

Now, we bring all terms to one side to form a standard quadratic equation:

Using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, where $a=1,b=1,c=-1$, we find the discriminant $D=1^2-4(1)(-1)=5$. Thus, the roots are:

The Reflection Path

Now, let us embrace the beauty of the even function property. We consider the case where $A=-B$. This means:

Again, we cross-multiply: $x(x+2)=-(x+1)$. Expanding this, we get $x^2+2x=-x-1$.

Rearranging the terms gives us a new quadratic equation:

Applying the quadratic formula again with $a=1,b=3,c=1$, the discriminant is $D=3^2-4(1)(1)=5$. The roots for this case are:

We have found four potential values for $x$: $\frac{-1\pm \sqrt{5}}{2}$ and $\frac{-3\pm \sqrt{5}}{2}$. But before we celebrate, we must act like true mathematicians and check our domain.

The function is defined on the interval $(-5,5)$. Since $\sqrt{5}\approx 2.23$, our roots are approximately $0.618,-1.618,-0.382,$ and $-2.618$.

All of these values fall comfortably within the interval $(-5,5)$. We have successfully navigated the trap and found all four solutions: