# inverse

• March 18th 2009, 11:56 AM
inverse
For what parameter m the function:
f(x) = (m+2)x^3-3mx^2+9mx-1 is invertible function.
• March 18th 2009, 12:43 PM
red_dog
f must be strictly increasing or decreasing.
That means $f'(x)>0$ or $f'(x)<0, \ \forall x\in\mathbf{R}$

$f'(x)=3[(m+2)x^2-2mx+3m]\Rightarrow (m+2)x^2-2mx+9m\neq 0, \ \forall x\in\mathbf{R}$

$\Delta =b^2-4ac=-8m(m+3)$

$\Delta<0\Rightarrow m\in(-\infty,-3)\cup(0,\infty)$

In this case f is injective.

If $m\in(-\infty,-3)\Rightarrow\lim_{x\to -\infty}f(x)=\infty, \ \lim_{x\to\infty}f(x)=-\infty$

If $m\in(0,\infty)\Rightarrow\lim_{x\to-\infty}f(x)=-\infty, \ \lim_{x\to\infty}f(x)=\infty$

In both cases f is continuous, so $f(\mathbf{R})=\mathbf{R}$ and f is surjective.