/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 Show that \(2\left(\frac{e^{x}-e... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that \(2\left(\frac{e^{x}-e^{-x}}{2}\right)\left(\frac{e^{x}+e^{-x}}{2}\right)=\frac{e^{2 x}-e^{-2 x}}{2}\).

Short Answer

Expert verified
The LHS simplifies to \frac{e^{2x} - e^{-2x}}{2}, which matches the RHS.

Step by step solution

01

Simplify the Left-Hand Side (LHS)

Start with the expression given in the problem: \[ 2\frac{e^{x} - e^{-x}}{2} \frac{e^{x} + e^{-x}}{2} \] The factor 2 can be distributed into the fractions: \[ \frac{e^{x} - e^{-x}}{1} \frac{e^{x} + e^{-x}}{2} \]
02

Multiply the Numerators

Combine and multiply the numerators: \[ (e^x - e^{-x})(e^x + e^{-x}) \] Using the identity for the difference of squares, this becomes: \[ e^{2x} - (e^{-x})^2 \] Simplify it further to: \[ e^{2x} - e^{-2x} \]
03

Simplify the Denominator

Combine and simplify the denominators as follows: \[ e^{2x} - e^{-2x} \] The denominator remains 2: \[ \frac{e^{2x} - e^{-2x}}{2} \]
04

Conclusion

The left-hand side (LHS) and the right-hand side (RHS) of the equation both simplify to the same expression: \[ \frac{e^{2x} - e^{-2x}}{2} \]. Therefore, the original equation is proved to be true.

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Key Concepts

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

Difference of Squares
The difference of squares is a fundamental algebraic identity that can help simplify complex expressions. It states that for any two quantities, say \(a\) and \(b\), the expression \(a^2 - b^2\) can be factored as \((a - b)(a + b)\). This identity helps in breaking down polynomial expressions into simpler linear factors. For example, consider the expression \((e^x - e^{-x})(e^x + e^{-x})\). Here, \(a = e^x\) and \(b = e^{-x}\). Applying the difference of squares, we get:
\
\[ (e^x - e^{-x})(e^x + e^{-x}) = e^{2x} - (e^{-x})^2 \]
Simplifying further, we get:
\
\[ e^{2x} - e^{-2x} \]
Using this identity can transform a seemingly complicated expression into a much simpler one, allowing for easier manipulation and calculation.
Exponential Functions
Exponential functions are a type of mathematical function where the variable is in the exponent. Typical notations involve bases like \(e\), which is approximately equal to 2.718. When dealing with exponential functions, it's important to understand the rules of exponents and how they interact:
  • For any number \(a\), \(a^m \times a^n = a^{m+n}\)
  • Similarly, \(\frac{a^m}{a^n} = a^{m-n}\)
  • And, \((a^m)^n = a^{mn}\)
  • Especially for base \(e\), \(e^0 = 1\) and \(e^1 = e\)
In our exercise, we see exponential terms such as \(e^x\) and \(e^{-x}\). While working on the expression \((e^x - e^{-x})(e^x + e^{-x})\), understanding these properties allows us to simplify them using algebraic identities like the difference of squares. Recognizing how to manipulate these terms helps divide, multiply, and factor exponential expressions efficiently. Hence, mastering exponential functions is crucial for simplifying and solving many mathematical problems.
Simplifying Expressions
Simplifying expressions involves reducing them to their simplest form, making them easier to work with or solve. This can involve factoring, combining like terms, or using algebraic identities. Let's dive into the key steps to simplify complex expressions:
  • **Combine Like Terms:** Gather up and combine terms with the same variable and power.
  • **Factor GCF:** Factor out the greatest common factor (GCF) from terms in an expression.
  • **Use Identities:** Use algebraic identities, such as the difference of squares, to factorize and simplify.
  • **Simplify Fractions:** Reduce fractions to simplest form by finding and canceling out common factors from numerator and denominator.
In our example, we used the difference of squares and properties of exponents to simplify the expression:
\
\[ 2 \left(\frac{e^x - e^{-x}}{2}\right) \left(\frac{e^x + e^{-x}}{2}\right) \]
First, distribute the factor of 2:
\[ \frac{e^x - e^{-x}}{1} \frac{e^x + e^{-x}}{2} \]
Then, multiply the numerators and apply the difference of squares:
\[ (e^x - e^{-x}) (e^x + e^{-x}) = e^{2x} - e^{-2x} \] Finally, divide by 2:
\[ \frac{e^{2x} - e^{-2x}}{2} \]
Thus, the left-hand side simplifies to the right-hand side, proving the original expression. Simplifying expressions makes mathematical operations more manageable and helps in solving problems accurately and efficiently.

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Most popular questions from this chapter

A function of the form \(P(t)=a b^{t}\) represents the population of the given country \(t\) years after January 1,2000 . a. Write an equivalent function using base \(e\); that is, write a function of the form \(P(t)=P_{0} e^{k t} .\) Also, determine the population of each country for the year 2000 . $$\begin{array}{|l|c|c|c|} \hline \text { Country } & P(t)=a b^{t} & P(t)=P_{0} e^{k t} & \begin{array}{c} \text { Population } \\ \text { in } 2000 \end{array} \\ \hline \text { Haiti } & P(t)=8.5(1.0158)^{t} & & \\ \hline \text { Sweden } & P(t)=9.0(1.0048)^{t} & & \\ \hline \end{array}$$ b. The population of the two given countries is very close for the year 2000 , but their growth rates are different. Determine the year during which the population of each country will reach 10.5 million. c. Haiti had fewer people in the year 2000 than Sweden. Why did Haiti reach a population of 10.5 million sooner?

Solve for the indicated variable. \(A=P(1+r)^{t}\) for \(t\) (used in finance)

Solve the equation. Write the solution set with the exact values given in terms of common or natural logarithms. Also give approximate solutions to 4 decimal places. \(80=320 e^{-0.5 t}\)

Solve the equation. Write the solution set with the exact solutions. Also give approximate solutions to 4 decimal places if necessary. \(\log _{4}(3 w+11)=\log _{4}(3-w)\)

Two million \(E\). coli bacteria are present in a laboratory culture. An antibacterial agent is introduced and the population of bacteria \(P(t)\) decreases by half every \(6 \mathrm{hr}\). The population can be represented by \(P(t)=2,000,000\left(\frac{1}{2}\right)^{t / 6}\) a. Convert this to an exponential function using base \(e\). b. Verify that the original function and the result from part (a) yield the same result for \(P(0), P(6), P(12)\), and \(P(60) .\) (Note: There may be round- off error.)

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