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Chapter 0: Problem 2

For \(F(x)=x^{3}+3 x,\) find each value. (a) \(F(1)\) (b) \(F(\sqrt{2})\) (c) \(F\left(\frac{1}{4}\right)\) (d) \(F(1+h)\) (e) \(F(1+h)-F(1)\) (f) \(F(2+h)-F(2)\)

Short Answer

Expert verified
(a) 4, (b) 5鈭�2, (c) 49/64, (d) 4 + 6h + 3h^2 + h^3, (e) 6h + 3h^2 + h^3, (f) 15h + 6h^2 + h^3.

Step by step solution

01

Substitute and Simplify for F(1)

To find \( F(1) \), substitute \( x = 1 \) into the function.\[ F(1) = (1)^3 + 3 \times (1) \]Simplify the expression:\[ F(1) = 1 + 3 = 4 \]
02

Substitute and Simplify for F(\sqrt{2})

Substitute \( x = \sqrt{2} \) into the function.\[ F(\sqrt{2}) = (\sqrt{2})^3 + 3 \times \sqrt{2} \]Simplify the expression:\[ F(\sqrt{2}) = 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2} \]
03

Substitute and Simplify for F\left(\frac{1}{4}\right)

Substitute \( x = \frac{1}{4} \) into the function.\[ F\left(\frac{1}{4}\right) = \left(\frac{1}{4}\right)^3 + 3 \times \frac{1}{4} \]Simplify the expression:\[ F\left(\frac{1}{4}\right) = \frac{1}{64} + \frac{3}{4} \]Combine the fractions:\[ F\left(\frac{1}{4}\right) = \frac{1}{64} + \frac{48}{64} = \frac{49}{64} \]
04

Expand and Simplify for F(1+h)

Substitute \( x = 1 + h \) into the function.\[ F(1+h) = (1+h)^3 + 3(1+h) \]Expand \((1+h)^3\) using the binomial theorem:\[ (1+h)^3 = 1 + 3h + 3h^2 + h^3 \]Thus:\[ F(1+h) = 1 + 3h + 3h^2 + h^3 + 3 + 3h = 4 + 6h + 3h^2 + h^3 \]
05

Calculate F(1+h)-F(1)

To find \( F(1+h)-F(1) \), subtract \( F(1) \) (which is 4) from the expression for \( F(1+h) \):\[ F(1+h) - F(1) = (4 + 6h + 3h^2 + h^3) - 4 \]Simplify:\[ F(1+h) - F(1) = 6h + 3h^2 + h^3 \]
06

Expand and Simplify for F(2+h)

Substitute \( x = 2 + h \) into the function.\[ F(2+h) = (2+h)^3 + 3(2+h) \]Expand \((2+h)^3\):\[ (2+h)^3 = 8 + 12h + 6h^2 + h^3 \]Thus:\[ F(2+h) = 8 + 12h + 6h^2 + h^3 + 6 + 3h = 14 + 15h + 6h^2 + h^3 \]
07

Calculate F(2+h)-F(2)

To find \( F(2+h) - F(2) \), first find \( F(2) \):\[ F(2) = 2^3 + 3(2) = 8 + 6 = 14 \]Subtract \( F(2) \) from \( F(2+h) \):\[ F(2+h) - F(2) = (14 + 15h + 6h^2 + h^3) - 14 \]Simplify:\[ F(2+h) - F(2) = 15h + 6h^2 + h^3 \]

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Key Concepts

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

Function Evaluation
Function evaluation involves calculating the value of a function for particular values of its variable. In our example, the function is given as \( F(x) = x^3 + 3x \). When we want to find out the specific output of the function, we substitute different values into it.
  • For example, evaluating \( F(1) \) means substituting \( x = 1 \) into the function, which gives \( F(1) = 1^3 + 3 \times 1 = 4 \).
  • Similarly, \( F(\sqrt{2}) \) involves replacing \( x \) with \( \sqrt{2} \), resulting in \( F(\sqrt{2}) = (\sqrt{2})^3 + 3 \times \sqrt{2} = 5\sqrt{2} \).
Function evaluation helps us understand how the function behaves at different input values. It's the first and crucial step before proceeding with further algebraic manipulations.
Substitution Method
The substitution method is used to replace a variable in a function with a specific value. This method is fundamental in evaluating polynomial functions effectively. By substituting the given value into the function, we achieve the desired output directly.
  • For \( F\left(\frac{1}{4}\right) \), the substitution of \( x = \frac{1}{4} \) into the function results in computing \( \left(\frac{1}{4}\right)^3 + 3 \times \frac{1}{4} \), leading to \( \frac{49}{64} \).
  • In situations like \( F(1+h) \), we substitute \( x \) with \( 1+h \) to explore the function's behavior around 1. The expression simplifies by expanding \( (1+h)^3 \) and further adding the terms.
Substitution provides clarity on function changes when the variable takes on different forms or values.
Binomial Theorem
The binomial theorem is essential when expanding expressions involving powers, especially when the variable in the function is altered, such as in \( F(1+h) \) or \( F(2+h) \). The theorem allows us to expand expressions like \( (a + b)^n \) systematically.
  • For example, when expanding \( (1+h)^3 \), the binomial theorem gives us \( 1 + 3h + 3h^2 + h^3 \).
  • This expansion helps to derive the expression \( F(1+h) = 4 + 6h + 3h^2 + h^3 \) by ensuring all terms are accounted for correctly.
Using the binomial theorem simplifies complex polynomial expressions and helps in understanding their progression.
Simplification
Simplification is the process of reducing complex expressions into simpler forms, making them easier to interpret and use. In polynomial functions, simplification often follows evaluation, substitution, and expansion.
  • Consider the subtraction of two functions, like \( F(1+h) - F(1) \). First, substituting and expanding gives us a larger expression, which we then simplify to \( 6h + 3h^2 + h^3 \).
  • Simplification also involves combining like terms, handling fractions correctly, and reducing expressions to their most basic form without changing their value.
This process is crucial for solving equations efficiently and for verifying the correctness of the results obtained from function operations.

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Most popular questions from this chapter

The relationship between the unit price \(P\) (in cents) for a certain product and the demand \(D\) (in thousands of units) appears to satisfy $$ P=\sqrt{29-3 D+D^{2}} $$ On the other hand, the demand has risen over the \(t\) years since 1970 according to \(D=2+\sqrt{t}\) (a) Express \(P\) as a function of \(t\). (b) Evaluate \(P\) when \(t=15\).

Suppose that \(f\) is an even function satisfying \(f(x)=-1+\sqrt{x}\) for \(x \geq 0\). Sketch the graph of \(f\) for \(-4 \leq x \leq 4\)

Plot the graph of each equation. Begin by checking for symmetries and be sure to find all \(x\) - and \(y\) -intercepts. \(y=\frac{1}{x^{2}+1}\)

We now explore the relationship between \(A \sin (\omega t)+\) \(B \cos (\omega t)\) and \(C \sin (\omega t+\phi)\) (a) By expanding \(\sin (\omega t+\phi)\) using the sum of the angles formula, show that the two expressions are equivalent if \(A=C \cos \phi\) and \(B=C \sin \phi\) (b) Consequently, show that \(A^{2}+B^{2}=C^{2}\) and that \(\phi\) then satisfies the equation \(\tan \phi=\frac{B}{A}\). (c) Generalize your result to state a proposition about \(A_{1} \sin \left(\omega t+\phi_{1}\right)+A_{2} \sin \left(\omega t+\phi_{2}\right)+A_{3} \sin \left(\omega t+\phi_{3}\right)\) (d) Write an essay, in your own words, that expresses the importance of the identity between \(A \sin (\omega t)+B \cos (\omega t)\) and \(C \sin (\omega t+\phi) .\) Be sure to note that \(|C| \geq \max (|A|,|B|)\) and that the identity holds only when you are forming a linear combination (adding and/or subtracting multiples of single powers) of sine and cosine of the same frequency.

Prove that the operation of composition of functions is associative; that is, \(f_{1} \circ\left(f_{2} \circ f_{3}\right)=\left(f_{1} \circ f_{2}\right) \circ f_{3}\).
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