I understand how to solve quadratics, but I do not know how to approach this question. Could anyone show me a step by step solution expression $x$ in terms of $y$?
The explicit question out of the book is to find $f^{-1}(3)$ for $f(x) = x^5 +x^3+x$
So far I have reduced $x^5 +x^3+x = y$ to $y/x - 3/4 = (x^2 + 1/2)^2$ or $y = x((x^2+1/2)^2 + 3/4)$ but Im still just as lost.
On
Can you think of a value of $a$ so that $f(a) = 3$? Try plugging in some values of $a$ and see what you get.
If this value is unique (which is true, but we need to verify it), then we write $a = f^{-1} (3)$.
On
In order to find $f^{-1}(3)=x$ for some $x$. We can apply the function to both sides giving us $3=f(a)$ now we need to find $x^5+x^3+x=3\implies x^5+x^3+x-3=0$. There are many ways to solve this however you can go ahead and guess that one solution would be $x=1$. Now we can perform polynomial division on $x^5+x^3+x-3$ by dividing by $x-1$ as that is a root and we get $x^4+x^3+2x^2+3$. The rest of your roots are there however it seems they are not real.
There is almost an art to guessing however one way to get a good feel is to try and plot the graph for different values. I guessed 1 because there were 3 terms with no coefficients and they evaluate to 3.
On
How about we compute the inverse using the inverse function theorem?
The inverse function theorem tells us that if $f$ is differentiable and invertible, then $(f^{-1})'(f(x)) = \frac{1}{f'(x)}$. Your $f$ satisfies both those properties (as others have shown, it is strictly increasing and so injective and so invertible).
So we compute
$$ f'(x) = 5x^4 + 3x^2 + 1 $$
And we know that:
$$ (f^{-1})'(f(x)) = \frac{1}{5x^4 + 3x^2 + 1} $$
Finally, we also know two things. $f^{-1}$ is "the" anti-derivative of $(f^{-1})'$ (footnote 1) and
$$ \int (f^{-1})'(x)\, \mathrm{d}x $$
can be done by u-substitution.
But this is where I'm stopping, since I've been out of calculus for so long. :)
[1]: in this case, it matters that there isn't a unique anti-derivative. You will need to solve for the one with the correct constant $C$. And it is essential that you do so. Only the one with the correct constant is an inverse for $f$. The necessity to solve for the correct $C$ means that this method won't work on the general quintic, but it will work on this one. You do have enough information, since you can actually factor this quintic.
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To show that for any $b$ there is an $a$ such that $f(a)=b$ we can proceed like this.
The result is obvious if $b=0$. Now let $b\gt 0$. We show there is an $a$ such that $f(a)=b$.
Note that $f(0)\lt b$ and $f(b)\gt b$. So by the Intermediate Value Theorem, because $f$ is continuous, there is an $a$ between $0$ and $b$ such that $f(a)=b$.
A similar argument deals with negative $b$, or else we can use the fact that $f(-x)=-f(x)$
Uniqueness is dealt with in comments and other answers.
On
Taking into account the quality of previous answers and comments, my answer will probably be stupid.
If you want $x$ such that $$f(x)=x^5 +x^3+x - y=0$$ you could just apply iteratively Newton which will give as the simplest $$x_{n+1}=\frac{4 x_n^5+2 x_n^3+y}{5 x_n^4+3 x_n^2+1}$$ which could be considered as a recursive definition of the inverse of $y$.
Assuming $y>1$, starting with $x_0=y^{\frac{1}{5}}$ (since, for this value $f(x_0) >0$ and $f''(x_0)> 0$) the process will converge without any overshoot of the solution.
Let me try with $y=10^{10}$ so $x_0=100$; the successive iterates are $x_1=99.99799992$, $x_2=99.99799984$ which is the solution for ten significant figures.
Let me try with $y=10^{5}$ so $x_0=10$; the successive iterates are $x_1=9.979920876$, $x_2=9.979839999$ which is the solution for ten significant figures.
On
A few of the answers suggest guessing the answer by inspection. Indeed, in most calculus problems that could be on an exam it is very likely that the solution to the polynomial is meant to be 'obvious', frequently one of $0,\pm1,\pm2$. This is just to keep numbers and computations reasonable, but is not an absolute guarantee (especially if you get to use a calculator). The mantra is usually "the uglier it looks, the more likely it is that there is a simple answer and argument."
But there's also a formal justification. Our first hope would be for a rational solution to $f(x)=3$. Sounds like a lot of possibilities, right? But by the rational roots theorem the only possibilities here are $\pm1,\pm3$. That's a nice, small list, and it's easy to verify that one of these actually works, thus solving the problem.
If there wasn't a rational root we'd be stuck with a more problematic situation, but it's always a good idea to check for predictable, nice answers first, especially when it involves so little effort to do so.
Solving it for a general $y$ will be difficult, but solving for $y=0$ is more doable.
If $x^5+x^3+x=0$, then because the left hand side is divisible by $x$, either $x=0$ or we can divide by $x$ to get $(x^2)^2+x^2+1=0$. This is a quadratic in $x^2$, so by the quadratic formula, $x^2=\frac{-1\pm \sqrt{-3}}{2}.$
Thus, the five solutions are $x=0$ and $x=\pm\sqrt{\frac{-1\pm \sqrt{-3}}{2}}$.
If we want to show that the function is invertible, then because polynomial functions are continuous, it is enough to show that the function is either always increasing or always decreasing. Because polynomial functions are continuously differentiable and polynomials cannot be locally constant without being completely constant, this is equivalent to showing that the first derivative is always non-negative or always non-positive.
The derivative is $5(x^2)^2+3(x^2)+1$. Because $x^2\geq 0$ for all $x$, the derivative is always positive (in fact, always at least $1$), and therefore the function must be invertible.