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Chapter 2: Problem 52

Simplify the difference quotient \(\frac{f(x+h)-f(x)}{h}\) if \(h \neq 0\). $$f(x)=x^{3}-2$$

Short Answer

Expert verified
The simplified difference quotient is \(3x^2 + 3xh + h^2\).

Step by step solution

01

Write down the difference quotient

The difference quotient is given by \( \frac{f(x+h)-f(x)}{h} \). We will substitute \( f(x) = x^3 - 2 \) into this formula.
02

Compute \( f(x+h) \)

Find \( f(x+h) \) by substituting \( x+h \) into the function: \( f(x+h) = (x+h)^3 - 2 \).
03

Expand \( (x+h)^3 \)

Expand \( (x+h)^3 \) using the binomial theorem: \((x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \).
04

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

Substitute \( f(x+h) \) and \( f(x) \) into the difference: \(((x+h)^3 - 2) - (x^3 - 2) = x^3 + 3x^2h + 3xh^2 + h^3 - x^3 \). This simplifies to \( 3x^2h + 3xh^2 + h^3 \).
05

Divide by \( h \)

Substitute the expression obtained in Step 4 into the difference quotient and divide each term by \( h \): \( \frac{3x^2h + 3xh^2 + h^3}{h} = 3x^2 + 3xh + h^2 \).

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Key Concepts

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

Binomial Theorem
The binomial theorem is a powerful tool for expanding expressions that are raised to a power. In this particular exercise, we use it to expand \((x+h)^3\). The theorem states that for any positive integer \(n\), \((a+b)^n\) can be expanded as:
  • \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k\)
Here, we let \(a = x\) and \(b = h\), and \(n = 3\). By applying the binomial theorem, we expand \((x+h)^3\) to become:
  • \(x^3 + 3x^2h + 3xh^2 + h^3\)
Each term in the expansion combines powers of \(x\) and \(h\) with coefficients taken from Pascal's triangle. This process helps simplify the problem by turning a complex polynomial into a sum of simpler terms.
Polynomial Functions
Polynomial functions are expressions that involve variables raised to whole number powers. The most basic form of a polynomial function is \(f(x) = ax^n + bx^{n-1} + \cdots + z\), where each term is a product of a coefficient and a variable raised to a power.
In this exercise, the function \(f(x) = x^3 - 2\) is a cubic polynomial, meaning its highest degree is 3. Polynomial functions have distinct properties that make them straightforward to manipulate mathematically, such as:
  • Being smooth and continuous.
  • Having a finite number of roots.
  • Being easy to differentiate and integrate.
Understanding the structure and behavior of polynomial functions is crucial in calculus and precalculus, as they frequently appear in various mathematical problems.
Simplification of Expressions
Simplifying expressions is a fundamental skill in algebra and calculus. It involves reducing expressions to their simplest form. In this exercise, simplifying \(f(x+h) - f(x)\) means canceling common terms and factoring where possible.
Once we've expanded \((x+h)^3\) and substituted into the difference \(f(x+h) - f(x)\), we end up with \(x^3 + 3x^2h + 3xh^2 + h^3 - x^3\). Simplification occurs as follows:
  • Canceling out \(x^3\) since it's in both \(f(x+h)\) and \(f(x)\).
  • Combining the remaining terms: \(3x^2h + 3xh^2 + h^3\).
Ultimately, simplifying the expression makes it easier to further divide by \(h\) and approach the derivative.
Precalculus
Precalculus bridges the gap between algebra and calculus. It includes a variety of topics necessary for understanding calculus concepts, such as functions, complex numbers, and limits. In this exercise, understanding precalculus means being proficient with:
  • Manipulating functions like polynomials.
  • Expanding expressions accurately, such as using the binomial theorem.
  • Simplifying complex algebraic expressions to reveal underlying patterns.
These skills are foundational as they prepare students for studying calculus. By mastering these areas, students can tackle calculus problems with confidence, specifically those involving limits and differentiability, which are grounded in the algebraic manipulation studied in precalculus.

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Most popular questions from this chapter

Explain why the graph of the equation is not the graph of a function. $$x=-|y|$$

Algebraic methods were used to find solutions to each of the following equations. Now solve the equation graphically by assigning the expression on the left side to \(Y_{1}\) and the number on the right side to \(\mathbf{Y}_{2}\) and then finding the \(x\) -coordinates of all points of intersection of the two graphs. (a) \(x^{5 / 3}=32\) (b) \(x^{4 / 3}=16\) (c) \(x^{2 / 3}=-36\) (d) \(x^{144}=125\) (e) \(x^{1 / 2}=-27\)

Cars are crossing a bridge that is 1 mile long. Each car is 12 feet long and is required to stay a distance of at least \(d\) feet from the car in front of it (see figure). (a) Show that the largest number of cars that can be on the bridge at one time is \(15280 /(12+d)]\), where I I denotes the greatest integer function. (b) If the velocity of each car is \(v\) mi/hr, show that the maximum traffic flow rate \(F\) (in cars/hr) is given by \(F=[5280 v /(12+d)]\)

Estimate the solutions of the inequality. $$\left|1.2 x^{2}-10.8\right|>1.36 x+4.08$$

Sketch the graph of \(f\) $$f(x)=\left\\{\begin{array}{ll} x-3 & \text { if } x \leq-2 \\ -x^{2} & \text { if }-2
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