# How do you find the inverse of #f(x)= x^5+x^3+x#?

We can show that

Given:

We have:

We can find numerical approximations using Newton's method:

For an initial approximation we can use:

Then iterate to get better approximations using the formula:

graph{(y-x)(y-(x^5+x^3+x))(x-(y^5+y^3+y)) = 0 [-5.143, 4.857, -2.38, 2.62]}

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I think the inverse of this function is not expressible as an algebraic formula. You can prove that it exists and define:

Any other ideas?

In general quintic equations are not solvable by radicals.

Using Cardano's method, we can find that

You can define:

Update

Then you could express the solution in terms of the Bring radical.

In practice, this is way too cumbersome, does not really gain you much and you would be better off just approximating numerically.

In any case, this is beyond the scope of Precalculus level.

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To find the inverse of the function ( f(x) = x^5 + x^3 + x ), follow these steps:

- Replace ( f(x) ) with ( y ): ( y = x^5 + x^3 + x ).
- Swap the variables ( x ) and ( y ): ( x = y^5 + y^3 + y ).
- Solve the equation obtained in step 2 for ( y ).
- The resulting expression for ( y ) will be the inverse function of ( f(x) ).

However, finding an explicit algebraic expression for the inverse function of ( f(x) ) in this case might not be straightforward or possible. In some cases, it may be necessary to use numerical or graphical methods to approximate the inverse function.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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