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202205250911 Exercise 2.1A (Q1)

Exercises 1-8, show that the given function y=f(x) is one-to-one over the given interval, then find the formulae for the inverse function f^{-1} and its derivative.

1. f(x)=x\quad \forall\, x

2. f(x)=3x\quad \forall\, x
3. f(x)=x^2\quad \forall\, x\ge 0
4. f(x)=\sqrt{x}\quad \forall\, x\ge 0
5. f(x)=\displaystyle{\frac{1}{x}}\quad \forall\, x>0
6. f(x)=\displaystyle{\frac{1}{x}}\quad \forall\, x<0
7. f(x)=\displaystyle{\frac{1}{x^2}}\quad \forall\, x>0
8. f(x)=x^5\quad \forall\, x

Extracted from Michael Corral. (2020). Elementary Calculus.

Background. (one-to-one; inverse; derivative of an inverse)

A one-to-one function (aka injection/injective function) is a function f that maps distinct elements to distinct elements; i.e., f(x_1)=f(x_2)\Rightarrow x_1=x_2, or equivalently in the contrapositive that x_1\neq x_2\Rightarrow f(x_1)\neq f(x_2). Not to be confused with one-to-one correspondence that refers to bijection.

Wikipedia on Injective function

The inverse function of a function f (aka inverse of f) is a function that undoes the operation of f. The inverse of f exists if and only if f is bijective, and if it exists, is denoted by f^{-1}. For a function f:X\rightarrow Y, its inverse f^{-1}:Y\rightarrow X admits an explicit description: it sends each element y\in Y to the unique element x\in X such that f(x)=y.

Wikipedia on Inverse function

If y=f(x) is differentiable and has an inverse function x=f^{-1}(y), then f^{-1} is differentiable and its derivative is

\displaystyle{\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}}\quad\textrm{ if }\enspace\frac{\mathrm{d}y}{\mathrm{d}x}\neq 0.

Text on pg. 38, Sec. 2.1, Ch. 2

1.

f(x)=x\quad \forall\, x:

\begin{aligned} \forall\, x\in\mathbb{R},\quad & f(x_1)=x_1 \\ & f(x_2)=x_2 \\ \enspace & f(x_1)=f(x_2) \Rightarrow x_1=x_2 \\ \end{aligned}

\therefore y=f(x)=x is injective/one-to-one.

In order for an inverse f^{-1}(x) to exist, the original function f(x) must be bijective (et injective et surjective). It remains to check whether or not y=f(x)=x is surjective.

A function f:X\rightarrow Y is said to be surjective if \forall\, y\in Y,\enspace \exists\, x\in X\textrm{ s.t. }f(x)=y. In other words, every element of the function’s codomain is the image of at least one element of its domain.

Wikipedia on Surjective function

Assume f(x)=x is a real-valued function, i.e., f:\mathbb{R}\rightarrow\mathbb{R} given by x\mapsto x. Obviously f is surjective. Thus it is bijective enough to have an inverse f^{-1}:\mathbb{R}\rightarrow\mathbb{R}. And apparently f^{-1} is given by x\mapsto x, such that f^{-1}\circ f=f\circ f^{-1}=\textrm{id}_{\,\mathbb{R}}. Its derivative is \displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}\big( f^{-1}(x)\big) =\frac{\mathrm{d}}{\mathrm{d}x}(x)=1}.

The remaining questions are left the reader.