Is #f(x)=(x^3+2x^2-x-2)/(x+3)# increasing or decreasing at #x=-2#?

1 Answer
mason m
Dec 8, 2015

Increasing.

Explanation:

Find the first derivative at the given point. If the value is positive, the function is increasing. If the value is negative the function is decreasing.

Finding #f'(x)# will require the quotient rule:

#f'(x)=((x+3)d/dx[x^3+2x^2-x+2]-(x^3+2x^2-x-2)d/dx[x+3])/(x+3)^2#

#=((x+3)(3x^2+4x-1)-(x^3+2x^2-x-2))/(x+3)^2#

#=(3x^3+13x^2+11x-3-(x^3+2x^2-x-2))/(x+3)^2#

#=(2x^3+11x^2+12x-1)/(x+3)^2#

Now, calculate the derivative at the point.

#f'(-2)=(2(-8)+11(4)+12(-2)-1)/(-2+3)^2#

#=(-16+44-24-1)/1#

#=3#

Since #3>0#, the function is increasing when #x=-2#.

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