Prove

**chialin4** · February 14th 2010, 11:18 PM

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

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

**mathemagister** · February 15th 2010, 03:58 AM

Originally Posted by chialin4

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

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.

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

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

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

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

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:

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

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

Hope I helped

**chialin4** · February 15th 2010, 09:16 AM

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.

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

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

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

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

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:

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

This proves that $f(x)$ 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!!

**mathemagister** · February 15th 2010, 11:15 AM

Originally Posted by chialin4

thx for ur help!!!

such a nice guy!!

You're very welcome

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