Show that arctan(x)+e^x+x^3=0 has a unique solution.

Since either sketching the function f(x)=arctan(x)+e^x+x³ or evaluating the precise/approximated solutions of the equation would be impossible with A-level techniques, we have to come up with an "alternative method": the derivative one. First of all, we easily notice that the domain of the function is R and that it is continous on R (since it is a sum of continous functions). The derivative, which gives us the slope of the function, is f'(x)=1/(1+x²)+e^x+3x².
Now, 1/(1+x2)>0 for all x and so is ex. 3x2 is >=0 but when x=0 f(0)=2 so the derivative is always greater than 0. As a corollary of Lagrange's theorem, positive derivative implies strictly increasing function. Being f(x) continous and being the limit to -inf of f(x) = - inf and limit to +inf of f(x) = +inf, we can show that the function intersect the x-axis only once (Bolzano's theorem); therefore the given equation has a unique solution.