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

1 Answer
Aug 24, 2017

Answer:

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]}