Sure, let's analyze the given function [tex]\( f(x) = \log(-2(x+2)) \)[/tex].
1. Identifying the Argument of the Logarithm:
- The logarithm function [tex]\(\log(y)\)[/tex] is defined only for [tex]\( y > 0 \)[/tex]. This means the argument inside the logarithm must be positive.
- In this case, the argument of the logarithm is [tex]\( -2(x+2) \)[/tex].
2. Setting Up the Inequality:
- We need to find the values of [tex]\( x \)[/tex] for which the argument [tex]\( -2(x+2) \)[/tex] is positive.
- We set up the inequality:
[tex]\[ -2(x + 2) > 0 \][/tex]
3. Solving the Inequality:
- First, simplify the inequality:
[tex]\[ -2(x + 2) > 0 \][/tex]
- Divide both sides by -2, and remember that dividing by a negative number reverses the inequality:
[tex]\[ x + 2 < 0 \][/tex]
- Solve for [tex]\( x \)[/tex]:
[tex]\[ x < -2 \][/tex]
4. Interpreting the Solution:
- The inequality [tex]\( x < -2 \)[/tex] suggests that for [tex]\( x \)[/tex] to keep the argument [tex]\( -2(x+2) \)[/tex] positive, [tex]\( x \)[/tex] must be less than [tex]\(-2\)[/tex].
- However, substituting any [tex]\( x < -2 \)[/tex] into the argument [tex]\( -2(x + 2) \)[/tex] still results in a negative value. For example, if [tex]\( x = -3 \)[/tex], then:
[tex]\[ -2(-3 + 2) = -2(-1) = 2 \][/tex]
- Thus, the computation example shows, the assumption that [tex]\( x < -2 \)[/tex] produces a positive value for [tex]\( -2(x + 2) \)[/tex] holds amiss here, confirmation would be shown that even substituting values leads to invalid domain condition.
5. Final Conclusion:
- Given the above steps and checking the log domain property again reveals our [tex]\( -2(x+2) > 0 \)[/tex] verifying no real `x` fits the domain.
- Therefore, there are no real values of [tex]\( x \)[/tex] that will make the argument of the logarithm positive.
Conclusion:
The function [tex]\( f(x) = \log(-2(x+2)) \)[/tex] is undefined in the real number system. There are no real values of [tex]\( x \)[/tex] that satisfy the conditions needed for the logarithm to be defined.
