Sum of all x-coordinates

Algebra Level 3

$f(x)$ is a function defined at $x \in \mathbb R$ such that $f(1+x)=f(1-x)$ , and when $x \in [1,+\infty)$ :

$f(x)=\begin{cases} 1-|x-2| \ ,\ x \in [1,3) \\ 2f(\dfrac{x-1}{2}) \ ,\ x \in [3,+\infty) \end{cases}$

Let $g(x)=\begin{cases} \ln x \ ,\ x \in [1,\infty) \\ \ln(2-x) \ ,\ x \in (-\infty,1) \end{cases}$

Find the sum of x-coordinates of all the intersection points of $f(x)$ and $g(x)$ in the interval $[-5,7]$ .

6

5

9

7

1 solution

Sabhrant Sachan
Mar 19, 2020

$f(1+x) = f(1-x)$ , which means whatever value $f$ attains on moving $x$ distance from $1$ to the right is the same as moving $x$ distance from $1$ to the left , thus function $f(x)$ is symmetric about the line $x=1$ and simillary $g(x)$ is also symmetric about $x=1$ , as $g(1+h) = \ln(1+h)$ and $g(1-h) = \ln{(2-1+h)} = \ln{(1+h)}$ , where $h$ is any positive real number. This symmetry of $f$ and $g$ about $x=1$ line will help us in evaluating the sum of $x$ -coordinates.

We are looking for intersections in the interval $[1-6,1+6]$ . If $f$ and $g$ intersect at $x = 1+a$ then they will also intersect at $x = 1-a$ due to symmetry. Thus the sum of $x$ -coordinates will be $1 + 2\times$ (No. of times the graph intersect in $(1,7]$ ). By making the graph, we observe that they intersect $3$ times in the interval $(1,7]$ . The answer will be $7$ . An alternative method to calculate the answer will be appreaciated !

That’s how it should be done. It’s important to notice that y=lnx and y=x-1 has one and only one intersection point near x=1，which may easily fall to the fallacy that it has two intersection points.

Alice Smith - 1 year, 2 months ago

Sum of all x-coordinates

1 solution

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