Question

What is the range of the function `f(x)=|x^2-8x+7|`?

Answer 1

The range is: `0 <= f(x) < oo`

Explanation:

The quadratic `x^2 - 8x + 7` has zeros:

`x^2 - 8x + 7 = 0`

`(x-1)(x-7) = 0`

`x = 1 and x = 7`

Between 1 and 7 the quadratic is negative but the absolute value function will make these values positive, therefore, 0 is the minimum value of `f(x)`.

Because the value of the quadratic approaches `oo` as x approaches `+-oo`, the upper limit for f(x) does the same.

The range is `0 <= f(x) < oo`

Here is a graph of f(x):

graph{|x^2 - 8x + 7| [-15.04, 13.43, -5.14, 9.1]}