Is $f(x)=(-2x^3-6x^2-3x+2)/(2x-1)$ increasing or decreasing at $x=0$ ?

Question

1470 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-16T20:03:25

Decreasing

We think about evaluating $f(1)$ and $f(0)$
If $f(1)-f(0)<0$ then it is decreasing
if $f(1)-f(0)>0$ then it is increasing

Let's calculate $f(1)$ and $f(0)$

$f(1)=(-2(1)^3-6(1)^2-3(1)+2)/(2(1)-1)$

$f(1)=(-2-6-3+2)/(2-1)$
$f(1)=-9$

$f(0)=(-2(0)^3-6(0)^2-3(0)+2)/(2(0)-1)$
$f(0)=-2$

We have: $f(1)-f(0)=-9-(-2)=-9+2=-7$
So, $f(1)-f(0)<0$

$f(1)<f(0)$
So, the function is decreasing.

Is f(x)=(-2x^3-6x^2-3x+2)/(2x-1)f(x)=(-2x^3-6x^2-3x+2)/(2x-1) increasing or decreasing at x=0x=0?