How to find the distance between two funtion inputs given two function outputs

Question

The question:

Suppose you have a function given by $f(x)=0.1x^2-x+2.5$

We are given that

$f(x_1)=3$

$f(x_2)=1.5$

We are tasked with finding the distance between $x_1$ and $x_2$

My attempt so far:

I tried to find the solutions in each case using the quadratic formula, I was then going to subtract one x value from the x value found in the other solutions, but there are two problems come with doing that

1. There's two solutions for each value of $f(x)$ though by graphing we can figure out which would be reasonable

2. For the case where $f(x)=1.5$ we get imaginary solutions

My other thoughts :

I thought maybe something to do with finding the rate of change using differentiation and then noting that the distance between one of the $f$ values is half the other

But I don't that will work because it's a curve not a line and also I want to find a solution using more elementary methods . Like algebra.

Can anyone please help me out here?

Answer 1

You can find the points $x_1$ and $x_2$ using the quadratic formula. Given $f(x)=0.1x^2-x+2.5$ $$f(x_1) = 3 \Leftrightarrow 0.1x_1^2-x_1+2.5 = 3 \Leftrightarrow 0.1x_1^2 -x_1 -0.5 = 0$$ Multiplying both sides of the equation by $10$ we get $$x_1^2 - 10x_1 - 5 = 0 \Leftrightarrow x_1 = \frac{10 \pm \sqrt{10^2+20}}{2} = 5\pm\sqrt{30}$$ To find $x_2$ we do it the same way $$f(x_2) = 1.5 \Leftrightarrow 0.1x_2^2-x_2+2.5 = 1.5 \Leftrightarrow 0.1x_1^2 -x_1 + 1 = 0$$ Multiplying both sides of the equation by $10$ we get $$x_2^2 - 10x_2 +10 = 0 \Leftrightarrow x_1 = \frac{10 \pm \sqrt{10^2-40}}{2} = 5\pm\sqrt{15}$$

Now we can find the distance between then using than given two points $x,y\in \mathbb R$, $$d(x,y) = |x-y|$$

• $|5+\sqrt{30} - (5 +\sqrt{15})| = |5-\sqrt{30} - (5 -\sqrt{15})| = \sqrt{30} - \sqrt{15}$
• $|5+\sqrt{30} - (5 -\sqrt{15})| = |5-\sqrt{30} - (5 +\sqrt{15})| = \sqrt{30} + \sqrt{15}$