Solve Inverse Function: x^3 + x + x^(1/2)

[footmath]

please solve this inverse function :
$ x^{3}+x+x^{1/2}\ $

 

[DivisionByZro]

Can you please write it more clearly? What you have there isn't even a function of anything.

Perhaps you meant:

<br /> f(x)=x^{3}+x+x^{1/2}<br /> <br />

If so, what do you know about inverse functions? What is the general idea of finding an inverse function? Please show work before we help you any more.

 

[footmath]

I want to isolate expression for x

 

[HallsofIvy]

Let y= x^3+ x+ x^{1/2}. Make the substitution z= x^{1/2} so the first equation becomes y= z^6+ z^2+ z. Solving the first equation for x is equivalent to solving the equation z^6+ z^2+ z- y= 0 for z and then squaring. Since that last equation is a 6th degree polynomial equation, there will be no simple solution in terms of roots.

In, fact, strictly speaking, that function does not have a true inverse because it is not "one to one". Where did you get this problem?

 

[footmath]

For example f(x)=x^2 is not "one to one" but the inverse of is x^1/2 and in partular interval is acceptable.

 

[HallsofIvy]

For example f(x)=x^2 is not "one to one" but the inverse of is x^1/2 and in partular interval is acceptable.

I wouldn't put it that way. If f(x)= x^2 is defined on any interval of positive numbers, then its inverse is x^1/2. But if f(x)= x^2 is defined on an interval of negative numbers, its inverse is -x^1/2. Of course, if f(x)= x^2 on an interval that contains both positive and negative numbers, f has no inverse.

 

[spamiam]

I don't think there are any concerns about this function being one-to-one. For one thing, f(x) = x^3 + x +x^{1/2} is one-to-one since f&#039;(x) = 3x^2 + 1 + \frac{1}{2\sqrt{x}} is always positive. Also our original f is only defined on [0, \infty), so even after the substitution z = x^{1/2}, g(z) = z^6 + z^2 + z is still one-to-one on [0, \infty).

That said, finding the inverse still looks difficult. Is it possible to do something tricky like integrate the derivative of f^{-1}?

 

[JJacquelin]

That said, finding the inverse still looks difficult. Is it possible to do something tricky like integrate the derivative of f^(-1) ?

I don't thing so. The sextic equation cannot be analytically solved in terms of a finite number of elementary functions. Maybe with generalized hypergeometric functions.

 

[footmath]

How can solve this equation or isolate expression for x or inverse this function :
y=5x^6+3x^2+4
do you believe that this equation can not solve?

 

[JJacquelin]

Of course, y=5x^6+3x^2+4 can be solved because it is a particular case : Let X=x^2 and you can solve the resulting 3th degree equation.
But generally it isn't the case. for exemple y=x^6+x^2+x