How can I prove that exist exactly two points so that $f(x)=0$?

Question

given:

$$f(x)=\sin(x) +x^2-x-1$$

How can I prove that exist exactly two points so that $f(x)=0$ ?

I know that one of the points belongs to $[0,5]$ and the other to $[-\frac{\pi}{2},0]$ (by using in Intermediate value theorem) but I don't have idea how to prove that exists just two points and not more than to points..

Answer 1

Hint: If $f(x) = 0$ has more than two solutions, then what does that say about $f'(x) = 0$? And what does that say about $f''(x) = 0$?

Answer 2

Hint:

If you compute $f''(x)$, then, you'll notice that $f''(x)>0$ for all $x\in R$. This implies that the concavity of the function is the same throughout all $x$.

Now, if $f(x)$ had to have more than two roots, would it be possible? Consider the effect, of having $3$ roots, on the concavity of $f(x)$...

Answer 3

For fun:

Assuming $f''(x)$ exists , for $x$ real, and

$f''(x) \gt 0$, i.e the function is convex,

then $f$ has at most $2$ zeroes.

$f''(x) >0$ implies $f'(x)$ is strictly increasing.

Assume $f$ has more than $2$ distinct zeroes, e.g. $3$, or more:

By MVT there are $2$, or more, distinct $x_1,x_2$,..., such that

$f'(x_1)=f'(x_2) = ...= 0.$

A contradiction since f' is strictly increasing, it can cross the $X-$axis only once.