I think the title says it all; I’m looking for the inverse function of $\ x^2+x$, and I have no idea how to do it. I thought maybe you could use the quadratic equation or something. I would be interesting to know.
If you want to invert $x^2 + x$ on the interval $x \ge -1/2$, write $y = x^2 + x$, so $x^2 + x -y = 0$.
Use the quadratic equation with $a=1$, $b=1$, and $c=-y$ to find $$ x= \frac{-1 + \sqrt{1+4y}}{2}.$$
(The choice of whether to use $+\sqrt{4ac}$ rather than $-\sqrt{4ac}$ is because we are finding the inverse of the right side of the parabola. If you want to invert the left side, you would use the other sign.)
We can write $x^2+x=(x+\frac{1}{2})^2 – \frac{1}{4}$ by completing the square. If you restricted yourself to the domain $x \ge -\frac{1}{2}$ then this function would be invertible, and its inverse would be the function
$$x \mapsto \sqrt{x+\frac{1}{4}} – \frac{1}{2}$$
You could do a similar thing on the domain $x \le -\frac{1}{2}$; but the function has no well-defined inverse on $\mathbb{R}$.
Solve $y = x^2+x$, that is, solve the quadratic $x^2+x-y = 0$. This gives
$x = \frac{1}{2}(-1 \pm \sqrt{1+4y})$. Consequently there is an inverse iff $y \geq -\frac{1}{4}$.
To address issues raised in comments:
Let $f: \mathbb{R} \to \mathbb{R}$ be given by $f(x) = x^2+x$. Then since $f(-\frac{3}{2}) = f(\frac{1}{2}) = \frac{3}{4}$ is is clear that $f$ is not injective, furthermore, since $f(x) \geq -\frac{1}{4}$ it is clear that $f$ is not surjective, hence an inverse function, as usually defined, does not exist for $f: \mathbb{R} \to \mathbb{R}$.
This gives a few options:
(1) Define the function $\eta: \mathbb{R} \to 2^{\mathbb{R}}$ by $$ \eta(t) = \begin{cases} \{ \frac{1}{2}(-1 \pm \sqrt{1+4t})\}, && t \geq -\frac{1}{4} \\ \emptyset, && \text{otherwise}\end{cases}$$
Then we have $f(x) = y$ iff $x \in \eta(y)$. This was what I meant by the initial answer (with a lot less baggage).
(2) Restrict the domain of the original function and find a left inverse. For example, we could consider $f$ on the domain $[-\frac{1}{2},\infty)$, and the function $\phi :[-\frac{1}{4},\infty) \to \mathbb{R}$ by $\phi(t) = \frac{1}{2}(-1 + \sqrt{1+4t})$. Then we have $\phi \circ f = \text{Id}$ on the domain
$[-\frac{1}{2},\infty)$.
(3) Find a suitable right inverse and domain. For example, $f\circ \phi = \text{Id}$ on the domain $[-\frac{1}{4},\infty)$.
We have $x^2+x=y$ iff $x^2+x-y=0$. Solve this equation for $x$, using the Quadratic Formula. We get
$$x=\frac{-1\pm\sqrt{1+4y}}{2}.$$
Two answers (usually) for $x$, so we have a problem. It is fundamentally the same issue as the one we have with the simpler function $x^2$. And there is a very similar fix: restricting the domain of our original function.
Define a new function $f_{+}(x)$ by $f_{+}(x)=x^2+x$ if $x \ge -1/2$, with $f_{+}(x)$ undefined when $x\lt -1/2$. Then the function $g$ defined by
$$g(y)=\frac{-1+\sqrt{1+4y}}{2}$$
is the inverse function of $f_{+}$. Note that $g(y)$ is only defined when $y\ge -1/4$.
If we define the function $f_{-}(x)$ by $f_{-}(x)=x^2+x$ if $x\le -1/2$, with $f_{-}(x)$ undefined for $x\gt -1/2$, then the function $h$ defined by
$$h(y)=\frac{-1-\sqrt{1+4y}}{2}$$
is the inverse function of $f_{-}$. Note that $h(y)$ is only defined when $y\ge -1/4$.
Remark: It is useful to draw a picture to see what’s going on. The curve $y=x^2+x$ is an upward-facing parabola that meets the $x$ axis at $(-1,0)$ and $(0,1)$, and therefore has its vertex at $(-1/2, -1/4)$. For every $y$-value greater than $-1/4$, there are two $x$-values. If we want a function that “undoes” what $x^2+x$ did, which of these values will it pick?
Denote $f(x):=x^{2}+x$
Then $f$ is not invertible if you consider it as a function from$\mathbb{R}$
to $\mathbb{R}$ since, for example, $f(0)=f(-1)=0$ hence it is not
$1-1$
Hints:
See: http://www.wolframalpha.com/input/?i=Inverse%20Function%20x%5E2%2Bx&t=crmtb01 (look at the graphs
Can you follow these examples? http://tutorial.math.lamar.edu/Classes/Alg/InverseFunctions.aspx
Note: pay close attention to the range that your function is defined for as sometimes this cause no solution, so you have to limit the domain.
HTH ~A