How do you find the range of #f(x)=abs(x^2-8x+7)# for the domain #3 <= x <= 8#?

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Answer 1 · 2015-04-26T16:55:54

There are several approaches that will work quite well for this. Here's the one that I used:

Graph the parabola #y = x^2-8x+7#, the flip the negative part across the #x# axis to get the absolute value.

I chose this method, because it's relatively easy to find the #x# intercepts:
#x^2-8x+7 = 0#

#(x-7)(x-1) = 0#, the intercepts are #1,7# and the parabola opens upward:

The vertex has #x# coordinate equal to the midpoint between the intercepts, or #x=4# and #y# value
#y=(4)^2-8(4)+7 = 16-32+7 = -11#

graph{y = x^2-8x+7 [-18.32, 27.29, -11.85, 10.94]}

No graph the absolute values:

graph{y =abs( x^2-8x+7) [-7.87, 20.61, -1.88, 12.35]}

Now: #f(3) = 8# and #f(8) = 7#, but between #3# and #8# we will hit f(7)=0, so the range runs from #0# up to somewhere.

The maximum value of the absolute value on that domain will be the absolute value of the minimum of the parabola or #11#

The range for that domain is #0 <= y <= 11#