The following is Theorem 2.2.5 of Murphy's C*-algebras and operator theory:
Let $A$ be an unital C*-algebra and $a,b$ are positive invertable elements, if $a\leq b$, then $0\leq b^{-1}\leq a^{-1}$.
To show it, firstly the author claims if $c\geq 1$ then $c$ is invertable and $c^{-1}\leq 1$. While I think the following is a counterexample for this claim.
Let $A=C([2,4])$, then $f(x) = x+1 $ is a positive invertable function in $A$, and $f\geq 1$. Also $f^{-1}(x) = x-1$ but $f^{-1}\not \leq 1$.
Please tell me where my mistake is and also give me a proof for the claim. Thanks in advance.
For the proof of the author's claim, note that $c \geq 1 = c ^{-1} c$ if and only if $1 \geq c^{-1}$.
This example doesn't quite work since it doesn't meet the hypotheses set out by the author. Let's consider the C*-algebra $A := C([2,10])$ and the same function $f$, which you defined above. There are various notions of invertibility in the C*-algebra you're examining. In this setting, the operations are defined pointwise and so the inverse of your function should really be $1/(x+1)$, which does satisfy the required property.