1. Prove

show the function has an inverse
$\displaystyle f(x)=x^{3}-x+1$

the ft maybe do not have inverse
so
if have ,prove it!

my idea is use strictly insresing(decresing) to show
but i do not know the next step to show

2. Originally Posted by chialin4
show the function has an inverse
$\displaystyle f(x)=x^{3}-x+1$

the ft maybe do not have inverse
so
if have ,prove it!

my idea is use strictly insresing(decresing) to show
but i do not know the next step to show
Your idea would work fine. If you can prove that f(x) is always increasing or decreasing, you have proven that f(x) has an inverse.

$\displaystyle f(x)$ increases when $\displaystyle f'(x)>0$
$\displaystyle f(x)$ decreases when$\displaystyle f'(x)<0$

$\displaystyle f'(x) = 3x^2 -1$

Now we need to check if $\displaystyle f'(x)$ changes sign (crosses the x-axis). If it does change sign, the $\displaystyle f(x)$ does not have an inverse. First, we check for zeros of the derivative:

$\displaystyle 3x^2 - 1 = 0$
$\displaystyle x^2 = \frac{1}{3}$
Zeros of $\displaystyle f'(x)$ are:
$\displaystyle x=\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}$

This does not conclusively prove anything though, because it only says that the derivative has zeros, but not that it changes sign. To check if it changes sign, just figure out if the derivative is positive or negative in the regions between the zeros:

$\displaystyle f'(0)=-1$ [substitute 0 into the derivative equation] Since the derivative is negative, $\displaystyle f(x)$ is decreasing at x=0.
$\displaystyle f'(1)=2$ Since the derivative is positive, $\displaystyle f(x)$ is increasing at x=1.

This proves that $\displaystyle f(x)$ does not have an inverse because it is increasing at some point, but decreasing at another.

Hope I helped

3. Originally Posted by mathemagister
Your idea would work fine. If you can prove that f(x) is always increasing or decreasing, you have proven that f(x) has an inverse.

$\displaystyle f(x)$ increases when $\displaystyle f'(x)>0$
$\displaystyle f(x)$ decreases when$\displaystyle f'(x)<0$

$\displaystyle f'(x) = 3x^2 -1$

Now we need to check if $\displaystyle f'(x)$ changes sign (crosses the x-axis). If it does change sign, the $\displaystyle f(x)$ does not have an inverse. First, we check for zeros of the derivative:

$\displaystyle 3x^2 - 1 = 0$
$\displaystyle x^2 = \frac{1}{3}$
Zeros of $\displaystyle f'(x)$ are:
$\displaystyle x=\frac{1}{\sqrt3}, -\frac{1}{\sqrt3}$

This does not conclusively prove anything though, because it only says that the derivative has zeros, but not that it changes sign. To check if it changes sign, just figure out if the derivative is positive or negative in the regions between the zeros:

$\displaystyle f'(0)=-1$ [substitute 0 into the derivative equation] Since the derivative is negative, $\displaystyle f(x)$ is decreasing at x=0.
$\displaystyle f'(1)=2$ Since the derivative is positive, $\displaystyle f(x)$ is increasing at x=1.

This proves that $\displaystyle f(x)$ does not have an inverse because it is increasing at some point, but decreasing at another.

Hope I helped

thx for ur help!!! such a nice guy!!

4. Originally Posted by chialin4
thx for ur help!!! such a nice guy!!
You're very welcome