Solutions to Problem Set 25

Math 310-01/02

11-17-2017

1. $f: \real \to \real$ is defined by $f(x) = x^{1/5} + 1$ . Prove that f is injective directly, using the definition of "injective".

Suppose $f(a) = f(b)$ . Then

$$\eqalign{a^{1/5} + 1 &= b^{1/5} + 1 \cr a^{1/5} &= b^{1/5} \cr (a^{1/5})^5 &= (b^{1/5})^5 \cr a &= b \cr}$$

Therefore, f is injective.


2. $f: \real \to \real$ is defined by $f(x) = x^5 + 2 x^3 + 10 x + 1$ . Prove that f is injective, using the derivative.

$f'(x) = 5 x^4 + 6 x^2 + 10 > 0$ for all x, so f is increasing for all x. Therefore, f is injective.


3. $f: \real \to \real$ is defined by $f(x) = x^{1/3} - 2$ . Prove that f is surjective directly, using the definition of "surjective".

Let $y \in \real$ . I need to find $x \in \real$ such that $f(x)
   = y$ .

First, I'll work backwards to "guess" the answer:

$$f(x) = y, \quad x^{1/3} - 2 = y, \quad, x^{1/3} = y + 2, \quad x = (y + 2)^3.$$

Next, I'll check that my guess works:

$$f\left((y + 2)^3\right) = \left((y + 2)^3\right)^{1/3} - 2 = y + 2 - 2 = y.$$

Therefore, f is surjective.


4. $f: \real \to \real$ is defined by $f(x) = e^x + x$ . Prove that f is surjective using the Intermediate Value Theorem.

Let $y \in \real$ . Note that

$$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (e^x + x) = +\infty \quad\hbox{and}\quad \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} (e^x + x) = -\infty.$$

Hence, there are numbers a and b such that

$$f(a) < y \quad\hbox{and}\quad y < f(b).$$

f is continuous, so by the Intermediate Value Theorem, there is an $x\in \real$ such that $f(x)
   = y$ .

Therefore, f is surjective.


Adversity makes no impression upon a brave soul. - Ali Ibn Abi Talib


Contact information

Bruce Ikenaga's Home Page

Copyright 2017 by Bruce Ikenaga