Textbook Question
Suppose ƒ is an even function with ƒ(2) = 2 and g is an odd function with g(2) = -2. Evaluate ƒ(-2) , ƒ(g(2)), and g(ƒ(-2))
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Ch. 1 - Functions
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 1, Problem 1.R.21
Evaluate and simplify the difference quotients (f(x + h) - f(x)) / h and (f(x) - f(a)) / (x - a) for each function.
f(x) = x2 - 2x
Verified step by step guidance1
Step 1: Identify the function f(x) = x^2 - 2x.
Step 2: For the first difference quotient, substitute x + h into the function: f(x + h) = (x + h)^2 - 2(x + h).
Step 3: Expand f(x + h): (x + h)^2 = x^2 + 2xh + h^2 and -2(x + h) = -2x - 2h, so f(x + h) = x^2 + 2xh + h^2 - 2x - 2h.
Step 4: Calculate the difference f(x + h) - f(x): (x^2 + 2xh + h^2 - 2x - 2h) - (x^2 - 2x) = 2xh + h^2 - 2h.
Step 5: Simplify the first difference quotient: (f(x + h) - f(x)) / h = (2xh + h^2 - 2h) / h = 2x + h - 2.
Step 6: For the second difference quotient, substitute a into the function: f(a) = a^2 - 2a.
Step 7: Calculate the difference f(x) - f(a): (x^2 - 2x) - (a^2 - 2a) = x^2 - 2x - a^2 + 2a.
Step 8: Simplify the second difference quotient: (f(x) - f(a)) / (x - a) = (x^2 - 2x - a^2 + 2a) / (x - a).
Step 9: Factor the numerator of the second difference quotient if possible to simplify further.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Difference Quotient
The difference quotient is a formula used to calculate the average rate of change of a function over an interval. It is expressed as (f(x + h) - f(x)) / h, where h represents a small change in x. This concept is fundamental in calculus as it leads to the definition of the derivative, which measures the instantaneous rate of change.
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The Quotient Rule
Derivative
The derivative of a function at a point quantifies how the function's output changes as its input changes. It is defined as the limit of the difference quotient as h approaches zero. Derivatives are essential for understanding the behavior of functions, including their slopes, rates of change, and optimization problems.
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Function Evaluation
Function evaluation involves substituting a specific value into a function to determine its output. For example, evaluating f(x) = x^2 - 2x at x = 3 involves calculating f(3) = 3^2 - 2(3). This concept is crucial for simplifying expressions in calculus, particularly when working with difference quotients and derivatives.
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Related Practice
Textbook Question
Simplify the difference quotient ƒ(x+h)-ƒ(x)/h
ƒ(x) = 2x² -3x +1
Textbook Question
Use the graph of ƒ to find ƒ⁻¹ (2),ƒ⁻¹ (9), and ƒ⁻¹ (12) <IMAGE>
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Textbook Question
Assume f is an odd function and that both f and g are one-to-one. Use the (incomplete) graph of f and the graph of g to find the following function values. <IMAGE>
f⁻¹( g⁻¹(4))
Textbook Question
Inverse sines and cosines Evaluate or simplify the following expressions without using a calculator.
sin⁻¹ ( -1 )
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Textbook Question
Assume f is an odd function and that both f and g are one-to-one. Use the (incomplete) graph of f and the graph of g to find the following function values. <IMAGE>
f⁻¹ (10)