Asked by MacDonald
Find all relative extrema and points of inflection for the function;
h(x)=(x^2+5x+4)/(x-1)
h(x)=(x^2+5x+4)/(x-1)
Answers
Answered by
Steve
h' = (x^2-2x-9)/(x-1)^2
h" = 20/(x-1)^3
Clearly there are no inflection points, since f" is never zero
h' is zero at two places, since the numerator is. So, there will be two extrema, one on each side of x=1. So, one will be a max, the other a min, depending on the sign of h".
Graph at:
http://www.wolframalpha.com/input/?i=+%28x^2%2B5x%2B4%29%2F%28x-1%29+
h" = 20/(x-1)^3
Clearly there are no inflection points, since f" is never zero
h' is zero at two places, since the numerator is. So, there will be two extrema, one on each side of x=1. So, one will be a max, the other a min, depending on the sign of h".
Graph at:
http://www.wolframalpha.com/input/?i=+%28x^2%2B5x%2B4%29%2F%28x-1%29+
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