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Chapter 0: Problem 82

\(f(x)=4 x^{2}-2 x, \quad \frac{f(x+h)-f(x)}{h}, h \neq 0\)

Short Answer

Expert verified
The result of the calculation with the difference quotient is \(8x-2\)

Step by step solution

01

Substitute the function into the formula

Substitute \(f(x+h)=4(x+h)^{2}-2(x+h)\) and \(f(x)= 4x^2-2x\) into the difference quotient formula: \[ \frac{(4(x+h)^{2}-2(x+h))-(4x^2-2x)}{h} \]
02

Simplify the expression

Do the arithmetic in the numerator: \[ \frac{4x^2+8xh+4h^2-2x-2h-4x^2+2x}{h} \]\n Simplify to get: \[ \frac{8xh+4h^2-2h}{h} \]
03

Divide by h

Finally, divide every term by \(h\): \[ 8x+4h-2 \]
04

Consider limit as h approaches 0

In the context where \(h\) is an infinitesimally small change, the \(4h\) term goes to zero, leaving: \(8x-2 \).

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Key Concepts

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

Function Substitution
When dealing with calculus problems, especially involving the difference quotient, one of the first techniques you'll often use is function substitution. This step is all about replacing variables or expressions to reformulate your equation.

In our original problem, this involved substituting the expression \( f(x+h) \) in place of \( f(x) \).
Here is how it works:
  • You have a function, in this case, \( f(x) = 4x^2 - 2x \).
  • When you substitute \( x + h \) for \( x \), you rewrite the function as \( f(x+h) = 4(x+h)^2 - 2(x+h) \).

This allows you to form the difference quotient, a key step in finding derivatives by capturing the rate of change over a small interval \( h \).
Applying substitution may look simple, but it lays the foundation for developing more advanced calculus skills.
Simplifying Expressions
Next up, simplifying expressions is an essential algebra skill that you will use extensively in calculus.
In our step-by-step solution, simplification involved organizing and reducing a complex expression into a simpler form.

Once we've substituted and plugged \( f(x+h) \) and \( f(x) \) into our difference quotient formula, the next task is to simplify the numerator:
  • The expression starts as \( 4(x+h)^2 - 2(x+h) - (4x^2 - 2x) \).
  • You expand, combine like terms, and aim to cancel out unnecessary components, resulting in \( 8xh + 4h^2 - 2h \).

This simplification is crucial because it prepares the expression for the next steps, like dividing by \( h \), and sets up for more advanced manipulations in calculus.
Keeping expressions simple makes them easier to handle and helps in avoiding mistakes.
Limits in Calculus
One of the cornerstones of calculus is the concept of limits, which you will find particularly useful when working with difference quotients. Limits help us understand behavior as a variable approaches a certain value.
In our problem, we use limits to determine what happens as the interval \( h \) gets closer and closer to zero.

Here's how limits come into play:
  • After substituting and simplifying, you have \( \frac{8xh + 4h^2 - 2h}{h} \).
  • Dividing each term by \( h \) results in \( 8x + 4h - 2 \).
  • As \( h \) approaches zero, the term \( 4h \) becomes negligible.
  • The limit, therefore, simplifies the expression further to \( 8x - 2 \), which is the derivative of the original function \( f(x) \).

Understanding limits is key to grasping how calculus allows us to find instantaneous rates of change, a concept central to derivatives and integrals.

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Most popular questions from this chapter

Use a graphing utility to graph each function. Write a paragraph describing any similarities and differences you observe among the graphs. (a) \(y=x\) (b) \(y=x^{2}\) (c) \(y=x^{3}\) (d) \(y=x^{4}\) (e) \(y=x^{5}\) (f) \(y=x^{6}\)

The set of ordered pairs \(\\{(-8,-2),(-6,0),(-4,0)\), \((-2,2),(0,4),(2,-2)\\}\) represents a function.

Determine whether the function is even, odd, or neither. Then describe the symmetry. $$ h(x)=x^{3}-5 $$

Geometry A rectangle is bounded by the \(x\)-axis and the semicircle \(y=\sqrt{36-x^{2}}\) (see figure). Write the area \(A\) of the rectangle as a function of \(x\), and determine the domain of the function.

In Exercises 33-38, use a graphing utility to graph the function, and use the Horizontal Line Test to determine whether the function is one-to-one and so has an inverse function. $$ g(x)=\frac{4-x}{6} $$
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