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Chapter 6: Problem 126

If \(f(x)=a^{x},\) show that $$ \frac{f(x+h)-f(x)}{h}=a^{x} \cdot \frac{a^{h}-1}{h} \quad h \neq 0 $$

Short Answer

Expert verified
\( \frac{f(x+h) - f(x)}{h} = a^x \cdot \frac{a^h - 1}{h} \)

Step by step solution

01

Write the Definition

Start with the given function definition: \(f(x) = a^x\). Consider the expression we need to prove: \( \frac{f(x+h) - f(x)}{h} \).
02

Substitute \(f(x)\) and \(f(x+h)\)

Substitute the terms in the expression with the function definition: \( \frac{a^{x+h} - a^x}{h} \).
03

Utilize Exponent Properties

Rewrite \(a^{x+h}\) as \(a^x \times a^h\). Thus, the expression becomes \( \frac{a^x \times a^h - a^x}{h} \).
04

Factor Out \(a^x\)

Factor out \(a^x\) from the numerator: \( \frac{a^x (a^h - 1)}{h} \).
05

Simplify the Expression

Separate \(a^x\) and the remaining fraction: \( a^x \times \frac{a^h - 1}{h} \).
06

Observe the Final Expression

The obtained expression matches the target: \( \frac{f(x+h) - f(x)}{h} = a^x \cdot \frac{a^h - 1}{h} \). This completes the proof.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Functions
Exponential functions are a type of mathematical function where a constant base is raised to a variable exponent, represented as \(f(x) = a^x\). Here, 'a' is a constant, and 'x' is the variable exponent. Exponential functions grow faster than polynomial functions.

They have several key properties:
  • The base, 'a', must be a positive real number and not equal to 1.
  • The function is always positive for all real numbers.
  • Exponential growth occurs when the base is greater than 1; decay happens when the base is between 0 and 1.
Understanding the structure and behavior of exponential functions is crucial before diving into their derivatives.
Derivatives
The derivative of a function measures how the function's output value changes as its input value changes. It's a fundamental concept in calculus and provides the rate of change or the slope of the function at any given point.

For exponential functions, this involves:
  • Applying the limit definition of the derivative, which is given by \(\lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}\).
  • Analyzing how small changes in the input (x) affect the output (\(f(x)\)).
In our exercise, we use this definition to find the derivative of the exponential function \(f(x) = a^x\).
Algebraic Manipulation
Algebraic manipulation involves the process of rearranging and simplifying mathematical expressions to obtain a desired form. This skill is essential in calculus, especially when working with derivatives and integrals.

In the given exercise:
  • We started by substituting the exponential functions into the derivative's limit definition formula.
  • We used properties of exponents to break down and simplify the expression (for instance, \(a^{x+h} = a^x \cdot a^h\)).
  • We factored out common terms to make the equation more manageable.
These steps showcase common techniques in algebraic manipulation.
Limits
Limits are a foundational concept in calculus, describing the value that a function approaches as the input approaches some value. They are essential for defining derivatives and integrals.

In our context:
  • We examined the behavior of an expression as h approaches 0, which is a standard approach for finding derivatives.
  • The formula \(\lim_{{h \to 0}} \frac{a^h - 1}{h}\) is particularly important for understanding the rate of change in exponential functions.
  • Using limits helps us precisely determine the derivative of the exponential function, confirming that \(\frac{f(x+h) - f(x)}{h} = a^x \cdot \frac{a^h - 1}{h}\) for non-zero h.
Grasping how limits work is crucial for mastering more advanced calculus topics.

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