How to prove that the function $x^2 + x \cos(x)$, $x \geq 0$ is one-to-one?

Question

I want to prove that the function $f(x)= x^2 + x \cos(x) $ is one-to-one, for $x \geq 0$.

1. First approach: the derivative of the function is $$f'(x) = 2x + \cos(x) - x\sin(x) = \cos(x) + x(2-\sin(x)).$$

Now, I know that $(2-\sin(x)) \geq 1 $, this means that $f'(x) \geq \cos(x) + x$.

When I plot $\cos(x) + x$, I see it is positive for $ x \geq 0$. How can I prove this mathematically?

2. Second approach: Let $x_1, x_2 \geq 0$ and assume that $f(x_1) = f(x_2)$. Then $$ x_1^2 + x_1 \cos(x_1) = x_2^2 + x_2 \cos(x_2) $$ $$ \ldots ?$$ $$ x_1 = x_2$$

Answer 1

$$f(x) = x + cos(x)$$ $$f'(x) = 1-sin(x) \geq 0$$ Hence $f(x)$ is always non decreasing, so minimum of $f(x)$ occurs at $x=0$ Since $f(0)=1$, we have $f(x) \geq 1, x \geq 0 $

Hence your original function is one-to-one

Answer 2

Consider $g(x) = \cos x + x$. $g^\prime(x) = - \sin x + 1 \ge 0$. Hence $g$ is stricly increasing for $x \in \mathbb R \setminus \{\pi/2 + 2k\pi \mid k \in \mathbb Z\}$. As $g(0) = 1$, $f^\prime$ is stricly positive for $x \ge 0$.