Why $ g(p) = 0.5 p^{-0.2} + 0.5 p^{-0.5} $ has a well-defined inverse that is continuous and strictly decreasing. A book that I am reading claims the following about the function $ g(p) = 0.5 p^{-0.2} + 0.5 p^{-0.5} $ (which is a demand function):

Formal arguments based on the Intermediate Value Theorem and the Implicit Function
    Theorem would establish that the inverse demand function is well-defined, continuous
    and strictly decreasing.

I have done some reading on the Intermediate Value Theorem and the Implicit Function Theorem. I understand what the Intermediate Value Theorem is about, but I am not so sure if I have understood the Implicit Function Theorem.
My question is: What might be those formal arguments based on these two theorems, and how do they show that the inverse function is well-defined, continuous and strictly decreasing?
 A: For all $ p \in (0,\infty) $, observe that we have $ \dfrac{0.5}{\sqrt[5]{p}} + \dfrac{0.5}{\sqrt{p}} \in (0,\infty) $. We can thus formally define a function $ g: (0,\infty) \to (0,\infty) $ by
$$
\forall p \in (0,\infty): \quad g(p) \stackrel{\text{def}}{=}
\frac{0.5}{\sqrt[5]{p}} + \frac{0.5}{\sqrt{p}}.
$$

Elucidating the properties of $ g $


*

*$ g $ is continuous on $ (0,\infty) $, as it is a sum of continuous functions from $
  (0,\infty) $ to $ (0,\infty) $.

*$ g $ is strictly decreasing on $ (0,\infty) $, as it is a sum of strictly decreasing
functions from $ (0,\infty) $ to $ (0,\infty) $. This proves that $ g $ is $ 1 $-$ 1
  $.

*Note that $ \displaystyle \lim_{p \to 0^{+}} g(p) = \infty $ and $ \displaystyle
  \lim_{p \to \infty} g(p) = 0 $. Hence, by the Intermediate Value Theorem, for every $
  y \in (0,\infty) $, we can find a $ p \in (0,\infty) $ such that $ y = g(p) $. This
proves that $ g $ is onto.

*As $ g $ is $ 1 $-$ 1 $ and onto, $ g^{-1}: (0,\infty) \to (0,\infty) $ exists.

*$ \color{darkgreen}{\text{Proof that $ g^{-1} $ is strictly decreasing:}} $

Pick $ y_{1} < y_{2} $ in $ (0,\infty) $. As $ g $ is strictly decreasing and
  $$
    g({g^{-1}}(y_{1})) = y_{1} < y_{2} = g({g^{-1}}(y_{2})),
    $$
  it follows that $ {g^{-1}}(y_{1}) > {g^{-1}}(y_{2}) $.


*$ \color{darkgreen}{\text{Proof that $ g^{-1} $ is continuous:}} $

As $ g^{-1} $ is strictly decreasing, the only discontinuities it can have are
  jump discontinuities. If $ g^{-1} $ had a jump discontinuity, then $ \text{Range}
    (g^{-1}) \subsetneq (0,\infty) $, which is clearly impossible as
  $$
    \text{Range}(g^{-1}) = \text{Domain}(g) = (0,\infty).
    $$

