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

1 Answer
Douglas K.
Aug 24, 2017

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]}

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