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

Question

1827 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-23T11:45:35

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

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!

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