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Chapter 3: Problem 43

Suppose \(f\) is a function with exponential growth. Show that there is a number \(b>1\) such that $$ f(x+1)=b f(x) $$ for every \(x\).

Short Answer

Expert verified
For a function with exponential growth \(f(x) = ab^x\), with \(a \neq 0\) and \(b > 1\), evaluating and simplifying \(f(x+1)\) reveals that \(f(x+1) = ab^{x} * b = f(x) * b\). Therefore, there exists a number \(b > 1\) that satisfies the equation \(f(x+1) = bf(x)\) for every \(x\), as required.

Step by step solution

01

Understanding exponential growth

An exponential growth function is a function which grows exponentially with increasing values of \(x\). Mathematically, this means that for some constant \(a \neq 0\) and \(b > 1\), we have the function in the form: \(f(x) = ab^x\) It is essential to recall that the base, \(b\), in the function form must be greater than 1 for it to represent exponential growth.
02

Evaluate \(f(x+1)\)

Now, let's evaluate the function at \(f(x+1)\) using the exponential growth function: \(f(x+1) = ab^{x+1}\)
03

Simplify \(f(x+1)\)

Next, we can simplify the expression for \(f(x+1)\) using the properties of exponents: \(f(x+1) = ab^{x} * b\)
04

Relate \(f(x+1)\) to \(f(x)\)

Now, let's look at the function form for \(f(x)\): \(f(x) = ab^x\) We can notice that an expression for \(f(x+1)\) is related to \(f(x)\) as follows: \(f(x+1) = f(x) * b\) As we can see, there exists a number \(b > 1\) that satisfies the equation \(f(x+1) = bf(x)\) for every \(x\). Therefore, we have proven that for a function with exponential growth, there is a number \(b>1\) such that \(f(x+1)=bf(x)\) for every \(x\).

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