Question

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

Answer 1

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

Explanation:

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!