# How do you find the point c in the interval 1<=x<=4 such that f(c) is equation to the average value of f(x)=abs(x-3)?

Mar 31, 2018

$x \in \left\{\frac{13}{6} , \frac{23}{6}\right\}$

#### Explanation:

First, let's find the average value:
$a v e \left(f \left(x\right)\right) = \frac{1}{4 - 1} {\int}_{1}^{4} f \left(x\right) \mathrm{dx}$

We think about the graph of $f \left(x\right)$ and see that the area is just two triangles:
graph{|x-3| [-1.085, 8.915, -0.68, 4.32]}

The leftmost triangle has base 3-1 = 2 and height of |1 - 3| = 2 so it has area 2.

The rightmost triangle has base 4 - 3 = 1 and height of |4 - 3| = 1, so it has area 1/2.

Therefore, the average value of the function is
$= \frac{1}{3} \cdot \left(2 + \frac{1}{2}\right) = \frac{5}{6}$

The function takes that value at two points:

$| x - 3 | = \frac{5}{6} \implies x - 3 = \pm \frac{5}{6} \implies x = 3 \pm \frac{5}{6}$

$x \in \left\{\frac{13}{6} , \frac{23}{6}\right\}$

As is hopefully clear in the above example, one uses calculus (here geometry) in order to find the average value and just algebra to find the values. We know that any continuous function has to cross its average value at least once within any interval, but there's no easy/formulaic way to find where that is true without doing algebra.