Problem 109 - College Algebra Essentials

Question

Find the domain of each logarithmic function. $$f(x)=\ln \left(x^{2}-x-2\right)$$

Step-by-Step Solution

Answer

The domain of the function is \(x < -1\) or \(x > 2\).

Step 1: Expression inside the Logarithmic Function

First step is to focus on the expression inside the logarithm, which is \(x^2-x-2\). In order to solve for \(x\), this quadratic equation must be factored and set to zero.

Step 2: Factoring

Looking at the quadratic expression, the factors of -2 that will add up to -1 are -2 and 1. So the factored form becomes \((x-2)(x+1) = 0\). Then, the solutions for \(x\) when this equation is equal to zero are \(x=2\) and \(x=-1\).

Step 3: The Interval of \(x\)

Keep in mind, logarithms are only defined for positive numbers, which means we're not interested in where the function equals zero, but where it is greater than zero. This notation is expressed as \(x^2 - x - 2 > 0\). To find where this is true, we need to examine the sign of the function within the intervals determined by the roots of the equation, in this case \(x < -1\), \(-1 < x < 2\), \(x > 2\).

Step 4: Evaluating the Quadratic Expression

Try a number from each interval into the factored form of the quadratic (from step 2), looking only at the sign \((x-2)(x+1) > 0\). For \(x < -1\), try \(x = -2\): \((-2 - 2)(-2 + 1) > 0\Rightarrow (-4)(-1) = 4 > 0\). So, the inequality holds true for \(x < -1\). For \(-1 < x < 2\), try \(x = 0\): \((0 - 2)(0 + 1) > 0\Rightarrow -2 < 0\). So, the inequality does not hold true for \(-1 < x < 2\). For \(x > 2\), try \(x = 3\): \((3 - 2)(3 + 1) > 0\Rightarrow 4 > 0\). So, the inequality holds true for \(x > 2\).

Step 5: Conclusion

So, the values for which the function is greater than zero are when \(x\) is less than -1 and when \(x\) is greater than 2. Thus the domain of the function \(f(x)=\ln (x^{2}-x-2)\) is \(x < -1\) or \(x > 2\).