Justify the solution of $x^2-x+\arctan{x}=0$ It is obvious that a solution of  $x^2-x+\arctan{x}=0$ is $x=0$, but I would like you to show me how this can be derived more conceptually than by plugging in $0$, and how can one prove that such solution is unique.
 A: Consider $f(x)=\arctan x$ and $g(x)=x-x^2$.
Both functions are increasing in $(-\infty,1/2]$, but the difference $f-g$ has a minimum in $x=0$, where $f(0)-g(0)=0$, so this the only point in which they have the same value.
After $x=1/2$ $f$ continue to increase while $g$ decreases, so no new contact could be present.
A: $0$ is an obvious root of the polynomial part, as it is lacking a constant term. $0$ is also an obvious root of the $arctan$ function, which is odd. And the sum of two terms sharing a root also has this root.
Now looking at the form of the expression, you notice that taking the first derivative will let the transcendence vanish: it is a rational fraction, creating hope that an exact study is feasible.
$$f'(x)=2x-1+\frac1{1+x^2}.$$
Indeed, the numerator is of the third degree so that a closed formula is available for the roots and you will be able to split the domain of $f$ into increasing and decreasing sections.
To complete the discussion, you will plug the root values in the expression of $f$ and evaluate their signs.
In other, less easy cases you will have to replace the transcendental functions by more tractable bounds/approximations.
A: Hint: Show that the first derivative has constant sign to the left of 0 and constant sign to the right of 0.
A: Consider $f(x)=x^2-x+\arctan(x)$. Then $f'(x)=2x-1+\frac{1}{x^2+1}=\frac{2x^3-x^2+2x-1+1}{x^2+1}=x\frac{(2(x^2-1/2)^2+7/4)}{x^2+1}$.
Then the function $f$ is decreasing to the left of $x=0$ and increasing to the right.
A: Let $f(x)=x^2-x+\arctan(x)$.
Then $$f'(x)=2x-1+\frac{1}{x^2+1}=\frac{x\cdot(2x^2-x+2)}{x^2+1}$$ which has only $1$ real root $x = 0$. 
So when $x<0, \space f'(x)<0; \space x>0, \space f'(x)>0$.
At point $x = 0,\space f(x) = 0$, thus, there is no more such $x$, that $f(x) = 0$.
A: like you conjectured, $x = 0$ is the only solution.  i will first show that there are no positive solutions.
$$\arctan x = \int_0^x {dt \over 1 + t^2} = \int_0^x \left( 1 - t^2 + {t^4 \over 1 + t^2} \right) dt = x -{1 \over 3}x^3 + \int_0^x {t^4 \over 1 + t^4}dt$$
for $x > 0, \arctan x > x - {1 \over 3}x^3,$ therefore
$$x^2 - x + \arctan x > x^2 + {1 \over 3}x^3 > 0$$ the last inequality establishes the claim for $x > 0$
now for the case $x < 0$
$$\arctan x = \int_0^x {dt \over 1 + t^2} = \int_0^x \left( 1  - {t^2 \over 1 + t^2} \right) dt = x - \int_0^x {t^2 \over 1 + t^4}dt = x + \int_x^0 {t^2 \over 1 + t^4}dt$$
$$x^2 - x + \arctan x = x^2 + \int_x^0 {t^2 \over 1 + t^4}dt > x^2 > 0$$ the last inequality establishes the claim for $x < 0$
