Is #f(x)=(x-3)(x+15)(x+2)# increasing or decreasing at #x=-1#?

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Answer 1 · 2016-06-23T11:45:35

#f# is decreasing at #x=-1.#

For a fun. #f, f'(a)>0 rArr f# is incrs. at #a#.

For a fun. #f, f'(a)<0 rArr f# is dcrs. at #a#.

#f(x)=(x-3)(x+15)(x+2) =x^3+(-3+15+2)x^2+{-3*15+15*2+(-3)*2}x+(-3)*15*2#
#=x^3+14x^2-21x-90#.

Hence, #f'(x)=3x^2+28x-21 rArr f'(-1)=3-28-21=-46<0.#

So, #f# is decreasing at #x=-1.#

An easy way to find #f'(x)# is to use the Rule : #(uvw)'=u'vw+uv'w+uvw'.#

By this Rule, #f'(x)=(x+15)(x+2)+(x-3)(x+2)+(x-3)(x+15)# giving,
#f'(-1)=14*1+(-4)*1+(-4)14=14-4-56=-46,# as before!