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Chapter 1: Problem 10

Find \(f(x)\) at \(x=2\) if \(f(x)=\frac{(x+4)(x-8)}{(x+6)(x-6)}\)

Short Answer

Expert verified
So, \(f(2) = 1.125\).

Step by step solution

01

Check if the Point is in the Domain

Firstly, it is to take the denominator of the function \(f(x) = \frac{(x+4)(x-8)}{(x+6)(x-6)}\) and set it to 0 and simplify. This will give the points where the function is not defined. For the denominator \(x+6 = 0, x= -6\) and \(x-6 = 0, x=6\). Since \(x=2\) is not equal to -6 and 6, the function is defined at \(x=2\).
02

Substitute x in the Function

After confirming that \(x=2\) is within the domain of the function, substitute \(x=2\) into the function. Therefore \(f(2) = \frac{(2+4)(2-8)}{(2+6)(2-6)}\).
03

Simplify the Expression

Now, simplify the expression obtained. The result will be \(\frac{(6)(-6)}{(8)(-4)} = \frac{-36}{-32} = 1.125\).

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Key Concepts

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

Evaluating Functions
Evaluating functions is a fundamental process in calculus and algebra, enabling us to determine the value of a function at a specific point.
To evaluate a function, you simply substitute the given variable value into the function's expression.
  • For instance, if you have a function like \( f(x) = \frac{(x+4)(x-8)}{(x+6)(x-6)} \), and you want to find \( f(2) \), replace every \(x\) in the formula with 2.
  • Carrying out this substitution gives \( f(2) = \frac{(2+4)(2-8)}{(2+6)(2-6)} \).
  • Next, compute the arithmetic: \( 2+4 = 6 \), \( 2-8 = -6 \), \( 2+6 = 8 \), and \( 2-6 = -4 \).
This helps transform the rational expression into a straightforward arithmetic problem.
Rational Expressions
Rational expressions consist of one polynomial divided by another.
This is much like a fraction but with polynomials in both the numerator and/or the denominator.
In the function provided \( f(x) = \frac{(x+4)(x-8)}{(x+6)(x-6)} \), the numerator and denominator are products of linear expressions.
  • The numerator is \((x+4)(x-8)\), and it's a straightforward multiplication of two binomials.
  • The denominator \((x+6)(x-6)\) is approached in the same way.
When simplifying rational expressions, always aim to factorize and cancel common terms when possible.
Caution must be exercised to ensure the expression's domain is preserved during such simplifications.
Domain of a Function
The domain of a function refers to all possible input values (typically \(x\) values) that do not lead to undefined expressions.
For rational expressions, the domain is restricted by values that make the denominator zero.
In our example, the function \( f(x) = \frac{(x+4)(x-8)}{(x+6)(x-6)} \) has a domain defined by the exclusion of values that turn the denominator to zero.
  • Set each factor in the denominator equal to zero: \( x+6 = 0 \) and \( x-6 = 0 \) leading to the restrictions \( x eq -6 \) and \( x eq 6 \).
  • This means the function cannot take \(-6\) or \(6\) as an input value.
Thus, any \(x\) value apart from \(-6\) and \(6\) is in the domain, including \(x=2\), making it valid to evaluate \(f(x)\) at \(x=2\).

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Most popular questions from this chapter

The slope of the line tangent to the graph of \(3 x^{2}+5 \ln y=12\) at \((2,1)\) is (A) \(-\frac{12}{5}\) (B) \(\frac{12}{5}\) (C) \(\frac{5}{12}\) (D) \(-7\)

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