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Chapter 1: Problem 36

In Exercises \(29-44,\) find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) for each function. $$f(x)=-4 x^{2}+2 x-3$$

Short Answer

Expert verified
The difference quotient is \(-8x - 4h + 2\).

Step by step solution

01

Identify the Function

The given function is \( f(x) = -4x^2 + 2x - 3 \).
02

Substitute \(x + h\) into the Function

Calculate \( f(x+h) \) by substituting \( x + h \) into the function. This gives:\[f(x+h) = -4(x+h)^2 + 2(x+h) - 3.\]
03

Expand \((x + h)^2\)

Expand \((x + h)^2\) to \( x^2 + 2xh + h^2 \). Thus,\[f(x+h) = -4(x^2 + 2xh + h^2) + 2x + 2h - 3.\]
04

Simplify \(f(x + h)\)

Distribute \(-4\) across \(x^2 + 2xh + h^2\):\[f(x+h) = -4x^2 - 8xh - 4h^2 + 2x + 2h - 3.\]
05

Set Up the Difference Quotient

Substitute \(f(x)\) and \(f(x+h)\) into the difference quotient \(\frac{f(x+h) - f(x)}{h}\):\[\frac{-4x^2 - 8xh - 4h^2 + 2x + 2h - 3 - (-4x^2 + 2x - 3)}{h}.\]
06

Simplify the Numerator

Combine like terms in the numerator:\[(-4x^2 - 8xh - 4h^2 + 2x + 2h - 3) - (-4x^2 + 2x - 3)\]Simplifying gives:\[-8xh - 4h^2 + 2h.\]
07

Factor and Simplify the Difference Quotient

Factor an \(h\) out of the numerator:\[\frac{h(-8x - 4h + 2)}{h}.\]Cancel the \(h\) in the numerator and denominator to get:\[-8x - 4h + 2.\]
08

Conclusion

The difference quotient \(\frac{f(x+h) - f(x)}{h}\) simplifies to \(-8x - 4h + 2\).

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Key Concepts

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

Function Expansion
When you're working with functions like the one in the exercise, understanding function expansion is essential. Function expansion involves substituting a new expression into a function and distributing it to explore its new form. For example, our function here is a quadratic, represented as \( f(x) = -4x^2 + 2x - 3 \). The process of expanding becomes crucial when we need to substitute \( x+h \) into \( f(x) \).Here's what to do:
  • Substitute \( x + h \) for \( x \) in your original function, to get \( f(x+h) = -4(x+h)^2 + 2(x+h) - 3 \).
  • Next, expand \( (x+h)^2 \) by using the identity \((x+h)^2 = x^2 + 2xh + h^2\).
  • This results in a new expression that we can further simplify.
Understanding this step is crucial as it sets the stage for more complex operations and helps us derive the difference quotient.
Polynomial Functions
Polynomial functions, like the one given here, are expressions involving variables raised to whole number powers. They are found in many forms from simple linear expressions to complex quadratic ones like our function \( f(x) = -4x^2 + 2x - 3 \).Identifying elements in polynomial functions:
  • Coefficients: These are the numbers in front of each term. In \(-4x^2 + 2x - 3\), the coefficients are -4, 2, and -3.
  • Variable powers: The variable, usually denoted as \( x \), can have powers like \( x^2, x^3 \), etc. Here, we see \( x^2 \).
  • Constant term: A term without a variable, like \(-3\) in this function.
Polynomial functions are typically used to represent complex relationships and are central in calculus for deriving expressions like the difference quotient.
Simplifying Expressions
Simplifying expressions is a vital skill when working with mathematical functions, especially when calculating the difference quotient. Let's walk through how simplifying helps in the given exercise.After determining \( f(x+h) \) from the function expansion:
  • Ensure all like terms are combined. This may involve expanding the terms fully.
  • From \( -4(x^2 + 2xh + h^2) + 2x + 2h - 3 \), we distribute \(-4\) across each term to get \(-4x^2 - 8xh - 4h^2\).
  • Gather all similar terms which results in a neater, simplified form \(-8xh - 4h^2 + 2h\).
  • Finally, factor out common terms to simplify further, as seen by factoring \( h \), leading to the expression \(-8x - 4h + 2\).
This final simplification is crucial to solve the difference quotient effectively and understand the behavior of the function as \( h \) approaches zero.

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Most popular questions from this chapter

Explain the mistake that is made. Given the function \(f(x)=\sqrt{x-2},\) find the inverse function \(f^{-1}(x),\) and state the domain restrictions on \(f^{-1}(x)\) Solution: Step 1: Let \(y=f(x)\). \(\quad y=\sqrt{x-2}\) Step 2: Interchange \(x\) and \(y . \quad x=\sqrt{y-2}\) Step 3: Solve for \(y\) \(y=x^{2}+2\) Step \(4: \operatorname{Let} f^{-1}(x)=y . \quad f^{-1}(x)=x^{2}+2\) Step 5: Domain restrictions \(f(x)=\sqrt{x-2}\) has the domain restriction that \(x \geq 2\) The inverse of \(f(x)=\sqrt{x-2}\) is \(f^{-1}(x)=x^{2}+2\) The domain of \(f^{-1}(x)\) is \(x \geq 2\) This is incorrect. What mistake was made?

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Refer to the following: In calculus, the difference quotient \(\frac{f(x+h)-f(x)}{h}\) of a function fis used to find the derivative \(f^{\prime}\) of \(f\), by allowing \(h\) to approach zero, \(h \rightarrow 0 .\) The derivative of the inverse function \(\left(f^{-1}\right)^{\prime}\) can be found using the formula $$\left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)}$$ provided that the denominator is not 0 and both \(f\) and \(f^{-1}\) are differentiable. For the following one-to-one function, find (a) \(f^{-1},\) (b) \(f^{\prime},\) (c) \(\left(f^{-1}\right)^{\prime},\) and (d) verify the formula above. For (b) and (c), use the difference quotient. $$f(x)=\sqrt{x+2}, x>-2$$

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In trigonometry you will learn about the tangent function, tan \(x\), Plot the function \(f(x)=\tan x,\) using a graphing utility. If you restrict the values of \(x\) so that \(-\frac{\pi}{2}< x< \frac{\pi}{2},\) the graph should resemble the graph below. Is the tangent function even, odd, or neither? GRAPH CAN'T COPY
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