Two questions on $f(x)=e^{-x}+\cos x$

Question

As mentioned, $f(x)=e^{-x}+\cos x$.

I have tried and having problems with:

1) Prove that $\inf f([0,\infty))=-1$ because it doesn't seem to be clear.
Do I calculate the limits in 0 and in $\infty^-$ and then I can decide which is the infimum? Can you show me how to do that?

2) Prove that for every $-1<c<2$ there is a solution to $f(x)=c$ in the range $[0,\infty]$.
How do I show that such a solution exists in that range? I tried to use the intermediate value theorem but didn't succeed. Can you show me how to do it? Or is there another way? I tried to show the continuity inside the given range to apply the intermediate value thorem, but I think it's wrong for some reason.

Thank you for your generous help.

Answer 1

We have

$$\forall x\ge 0 \;\;\; f (x)= e^{-x}+\cos (x)>-1$$

and

$$\lim_{n\to+\infty}f ((2n+1)\pi)=-1$$ thus $$\inf \{f (x),x\ge 0\}=-1$$

Answer 2

1) It is $f(x) = e^{-x} + \cos(x) \geq -1$ since $e^{-x} \geq 0$ and $\cos(x) \geq -1$. Further, $$ f((2n+1)\pi) = e^{-(2n+1)\pi} + cos((2n+1)\pi) = e^{-(2n+1)\pi} -1 \to -1$$ for $n \to \infty$, hence $\inf_{x \in [0, \infty)} f(x) = -1$.

2) Using the intermediate value theorem is the right idea. Since $f(0) = e^{0} + \cos(0) = 2$, by 1) $\inf f(x) = -1$ and $f$ is continuous, you can get the result you want.

Answer 3

As for the second question: yes, you apply the Intermediate Value Theorem, although with a twist. Notice that $f(0)=2$. Also, as $x\to+\infty$, we know that $e^{-x}\to0$ (from above), while $\cos x$ oscillates between $-1$ and $1$. In particular, $\cos x$ attains the value of $-1$ infinitely often as far away as we please. Pick some $c$ satisfying $-1<c<2$. Then we can pick a point $b$ far enough such that $f(b)$ is as close to $-1$ (from above) as we please, in particular we can make $-1<f(b)<c$. Applying the Intermediate Value on $[0,b]$ does the trick.

Answer 4

The second question isn't getting as much coverage as I think it should. The fact that the infimum of the function is $-1$ is indeed very helpful, but the function never attains the infimum of $-1$, so applying the Intermediate Value Theorem directly will not quite work.

Instead, let's suppose we're aiming for a function value $c$ between $-1$ and $2$. While $f(x) \neq -1$ for all $x$, because the infimum is $-1$, there must be some $x_0 \in \mathbb{R}$, such that $-1 < f(x_0) < c$, otherwise $c$ is a lower bound greater than the infimum!

Now, consider the Intermediate Value Theorem, between $x_0$ and $0$, and you'll get what you need.