Is #f(x)=(x^2-4x-9)/(x+1)# increasing or decreasing at #x=0#?

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Answer 1 · 2015-12-23T04:09:34

Increasing

Find #f'(0)#. If it's greater than #0#, the function is increasing at that point. If it's less than #0#, the function is decreasing at that point.

Use the quotient rule to find #f'(x)#.

#f'(x)=((x+1)d/dx[x^2-4x-9]-(x^2-4x-9)d/dx[x+1])/(x+1)^2#

#f'(x)=((x+1)(2x-4)-(x^2-4x-9)(1))/(x+1)^2#

#f'(x)=((2x^2-2x-4)-(x^2-4x-9))/(x+1)^2#

#f'(x)=(x^2+2x+5)/(x+1)^2#

#f'(0)=5#

Since #f'(0)>0#, #f# is increasing when #x=0#.

graph{(x^2-4x-9)/(x+1 [-44.5, 47.97, -27.72, 18.52]}