find number of solution of f(x)?

Question

$f(x)=(2\arctan(x))x + 1 - 3x$. I had to find the number of the solution of $f(x)$ -- just number of solutions, NOT SOLUTIONS TO EQUATION -- what I did was I derived f(x) then put it equal to 0 to find maxima and minima $f'(x)=0$ but I can't solve $f'(x)=0$. as $f'(x)=(2x)/(x^2 +1) +2 \arctan x - 3 = 0$ guys any help appreciated. or am I doing anything wrong

Answer 1

This we can also solve by drawing the graph of f(x) and checking how many times does it cut the x-axis.

• First, draw the graph for 2x*arctan(x) which will give you a parabola opened upward.
• Then subtract 1 from the graph which will give the vertex of the graph to be (0,1)
• Then add +3x to it, this will transform the graph as shown below, and the number of points where it intersects the x-axis will be the solution. Graph of the given function.

Here the solution is 2 as the graph intersects the x-axis two times.

Answer 2

The answer is to take the second derivative. We get $$f''(x)=\frac4{\left(1+x^2\right)^2}>0$$ Therefore, $f'$ is strictly increasing and $f'$ can have at most one zero, so $f$ has at most two zeros, by Rolle's theorem.

We know that $f(0) = 1$, and $$\lim_{x\to\infty}f(x) = \lim_{x\to\infty}x(2\arctan x -3)+1=\lim_{x\to\infty}x(\pi-3)+1 = \infty$$

Therefore, if we can find any $x>0$ such that $f(x)<0$ we know that $f$ has exactly two zeros, by the previous result and the Intermediate Value Theorem. Now, $$f(1)=2\arctan1-3+1=\frac\pi2-2<0$$ so $f$ has exactly $2$ zeros.