Can someone please help with steps? Thanks
f ' (x) = [absolute value x^2 - 4] / (x-2)
f is decreasing function on interval ?
f ' (x) = [absolute value x^2 - 4] / (x-2)
f is decreasing function on interval ?
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Given f ', we know that f is decreasing when f ' < 0. Simplify the given function.
|x² - 4| = |x - 2||x + 2|, so
f '(x) = |x - 2||x + 2|/(x - 2) = a |x + 2|
where a = 1 if x - 2 > 0 and a = -1 when x - 2 < 0. So f ' < 0 if x - 2 < 0 ----i.e. if x < 2. This is where f is decreasing.
Remember that |x - c|/(x - c) is 1 if x - c is positive and -1 if x - c is negative. It's not defined if x = c.
|x² - 4| = |x - 2||x + 2|, so
f '(x) = |x - 2||x + 2|/(x - 2) = a |x + 2|
where a = 1 if x - 2 > 0 and a = -1 when x - 2 < 0. So f ' < 0 if x - 2 < 0 ----i.e. if x < 2. This is where f is decreasing.
Remember that |x - c|/(x - c) is 1 if x - c is positive and -1 if x - c is negative. It's not defined if x = c.
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|x - 2| is always positive. x - 2 is negative when x < 2. So for x < 2, |x - 2|/(x - 2) is (2 - x)/(x -2) = -1.
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