Question

Suppose that \(f\) is defined on \((-\infty, \infty)\) and has the following properties. (i) \(\lim _{x \rightarrow 0} f(x)=1\) and (ii) \(\quad f\left(x_{1}+x_{2}\right)=f\left(x_{1}\right) f\left(x_{2}\right), \quad-\infty0\) for all \(x\). (b) \(f(r x)=[f(x)]^{r}\) if \(r\) is rational. (c) If \(f(1)=1\) then \(f\) is constant. (d) If \(f(1)=\rho>1,\) then \(f\) is increasing, $$ \lim _{x \rightarrow \infty} f(x)=\infty, \quad \text { and } \quad \lim _{x \rightarrow-\infty} f(x)=0 $$ (Thus, \(f(x)=e^{a x}\) has these properties if \(a>0 .\) )

Short answer

Question: Prove the following properties of the function \(f(x)\), given that \(f(x)\) is defined on \((-\infty, \infty)\), has a limit at 0, and satisfies the property \(f\left(x_{1}+x_{2}\right)=f\left(x_{1}\right) f\left(x_{2}\right)\): a) \(f(x)>0\) for all \(x\). b) \(f(r x)=[f(x)]^{r}\) if \(r\) is rational. c) If \(f(1)=1\), then \(f\) is constant. d) If \(f(1)=\rho>1\), then the function is increasing, with certain limits as \(x\) goes to \(\pm\infty\). Answer: We have proved that given the conditions of the function \(f(x)\), the following properties hold: a) \(f(x)>0\) for all \(x\), since the product of two non-zero numbers is positive only if both numbers are of the same sign. b) \(f(r x)=[f(x)]^{r}\) for any rational number \(r\), by showing that it holds for integers and extending it to rational numbers. c) If \(f(1)=1\), then the function is constant, as it does not change with \(x\). d) If \(f(1)=\rho>1\), then the function is increasing, with \(\lim_{x \rightarrow -\infty} f(x) = 0\) and \(\lim_{x \rightarrow \infty} f(x) = \infty\).

Step-by-step solution

Use the known properties of the function to show \(f(0)=1\)#

Since \(\lim_{x \rightarrow 0} f(x)=1\), we can deduce \(f(0)=1\). This is because a function is continuous at a point if the limit at that point exists and is equal to the function value at that point.

Prove \(f(-x) = \frac{1}{f(x)}\) for all \(x\)#

To prove this, we will use the property \(f\left(x_{1}+x_{2}\right)=f\left(x_{1}\right) f\left(x_{2}\right)\) and choose \(x_{1}=x\) and \(x_{2}=-x\). Thus: $$ f(x-x) = f(x)f(-x) $$ Since \(f(0)=1\), we get: $$ 1 = f(x)f(-x) $$ Hence, \(f(-x) = \frac{1}{f(x)}\) for all \(x\).

Show \(f(x)>0\) for all \(x\)#

Now, we can deduce that \(f(x)>0\) for all \(x\). We already know that \(f(0)=1>0\). For \(x>0\), we have \(f(x)f(-x)=1\). Since the product of two non-zero numbers is positive only if both numbers are of the same sign, this implies that \(f(x)>0\). Moreover, using the previous result, we have \(f(-x)=\frac{1}{f(x)}>0\). Hence, \(f(x)>0\) for all \(x\). To prove Property (b):

Prove \(f(r x)=[f(x)]^{r}\) for any rational number \(r\)#

First, we notice that \(f(n x)=[f(x)]^{n}\) for any integer \(n\) by applying the property of the function multiple times. For example, for \(n=3\), we have: $$ f(3x) = f(x+x+x) = f(x)f(x)f(x) = [f(x)]^3 $$ Now, let \(r=\frac{m}{n}\) be a rational number, with \(m\) and \(n\) being integers. By applying the property of the function: $$ f(m x) = [f(\frac{m}{n}x)]^n $$ And since \(f(m x) = [f(x)]^m\), we get: $$ [f(x)]^m = [f(\frac{m}{n}x)]^n $$ Hence, \(f(\frac{m}{n}x)= [f(x)]^{\frac{m}{n}}\), which implies \(f(r x)=[f(x)]^{r}\) for any rational number \(r\). To prove Property (c):

Show that if \(f(1)=1\), then \(f\) is constant#

In this case, we have: $$ f(x) = f(1\cdot x) = [f(1)]^xf(x)=1^xf(x)=f(x) $$ So, the function does not change with \(x\), and it is constant. To prove Property (d):

Show that if \(f(1)=\rho>1\), then \(f\) is increasing#

Let \(x>y\). Since \(f(1)=\rho>1\), and \(f\) is positive for all \(x\), we have: $$ f(x-y)=\frac{f(x)}{f(y)}>1 $$ Thus, \(f(x)>f(y)\), and the function is increasing.

Show the limits as \(x\) approaches \(\pm\infty\)#

An increasing function with a positive range has a finite limit when \(x \rightarrow -\infty\): $$ \lim_{x \rightarrow -\infty} f(x) = 0 $$ For the other limit, we have: $$ \lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} [f(1)]^x = \lim_{x \rightarrow \infty} \rho^x = \infty $$ So, we have proved all the properties, and the function can be characterized as \(f(x) = e^{ax}\) for \(a>0\).

Key concepts

Limits of Functions

The concept of a limit is foundational in real analysis and calculus. Limits help us analyze the behavior of functions as inputs approach a certain value. For example, when the exercise states \( \lim_{x \rightarrow 0} f(x)=1 \), it indicates that as the variable \(x\) gets closer to 0, the function's value approaches 1. This is crucial for understanding continuity, which requires that as \(x\) approaches a particular point, \(f(x)\) should approach \(f\) at that point. This foundational limit idea extends further to infinity, where we explore how a function behaves as \(x\) increases or decreases without bound.

Limits are not just about what happens exactly at a point, but rather what happens as we get close to it. They enable us to deal with situations where a function may not be defined at a point but has a predictable behavior as we approach that point.

Exponential Functions

Exponential functions represent an important class of functions characterized by a constant base raised to a variable exponent, typically expressed as \( f(x) = a^x \), where \( a \) is a positive constant. In our exercise, if we prove the function \([f(x) = e^{ax}]\) has a base \( \rho \), which is greater than 1, we see that \(f\) exhibits exponential growth. The exponential function is known for its unique property where the rate of change of the function is proportional to the value of the function itself.

Exponential growth is especially significant because it becomes more dramatic as \(x\) increases. The function doesn’t just increase, it increases at an ever-accelerating rate. This is reflected in our exercise, where we examine limits at infinity and determine that an increasing exponential function will tend toward infinity as \(x\) goes to infinity.

Continuous Functions

Continuity is a key concept in real analysis, describing functions that are unbroken or uninterrupted. A function \(f\) is continuous at a point \(x = a\) if \( \lim_{x \rightarrow a} f(x)=f(a) \). In other words, as \(x\) gets infinitesimally close to \(a\), the function's value approaches the actual value of \(f(a)\) without any jumps or gaps. This is closely tied to the notion of limits. The exercise showcases this fact through the function \( f \) which is implied to be continuous since it meets the criteria of the limit at \( x = 0 \) equalling the function value at the same point.

Being able to deduce that a function is continuous from its limit is invaluable. It provides a more comprehensive understanding of the function's behavior, assuring us that for every point within the function's domain, we can predict the function's behavior based solely on its limits.

Rational Powers

Rational powers extend the idea of exponentiation to fractions and involve roots as well as powers. When we raise a base \( x \) to a rational power \( r \), expressed as \( x^r \), we are in fact performing both an exponentiation and a root operation. For integer exponents, we multiply the base by itself repeatedly; for negative exponents, we take the reciprocal; and for fractional exponents, such as \( r = m/n \), we also involve the \( n \)th root.

In the context of our exercise, the ability to manipulate rational powers is essential to prove that the function \( f(rx)=[f(x)]^r \) for rational \( r \), which supports the broader understanding and properties of the function in question. This part of the exercise reveals the intrinsic connection between multiplication, powers, and roots, and how they operate within the realm of functions in real analysis.