Answer :
To determine on which interval both functions [tex]\( f(x) = x^2 + 2x \)[/tex] and [tex]\( g(x) = \log(2x + 1) \)[/tex] are negative, let's analyze each interval step by step:
1. Consider the interval [tex]\((-\infty, -2)\)[/tex]:
For [tex]\( f(x) = x^2 + 2x \)[/tex]:
- When [tex]\( x < -2 \)[/tex],
[tex]\( f(x) = x^2 + 2x \)[/tex] will be positive for [tex]\( x > -2 \)[/tex] and negative for [tex]\( x < -2 \)[/tex].
For [tex]\( g(x) = \log(2x + 1) \)[/tex]:
- The argument of the logarithm [tex]\( 2x + 1 \)[/tex] must be positive:
[tex]\[ 2x + 1 > 0 \implies x > -\frac{1}{2} \][/tex]
- However, since [tex]\( x < -2 \)[/tex], the argument [tex]\( 2x + 1 \)[/tex] will be less than [tex]\( 0 \)[/tex], making [tex]\( g(x) \)[/tex] undefined in this interval.
2. Consider the interval [tex]\((-\infty, -2)\)[/tex]:
For [tex]\( f(x) = x^2 + 2x \)[/tex]:
- When [tex]\( x < -2 \)[/tex],
[tex]\[ f(x) = x(x + 2) \][/tex]
- Since both [tex]\( x \)[/tex] and [tex]\( x+2 \)[/tex] are negative in the interval [tex]\((-\infty, -2) \)[/tex], [tex]\( f(x) \)[/tex] will be positive.
For [tex]\( g(x) = \log(2x + 1) \)[/tex]:
- The argument of [tex]\( g(x) \)[/tex] must be positive. For [tex]\( x < -2 \)[/tex], [tex]\( 2x + 1 < -3 \)[/tex] which makes the log function undefined or negative in the interval.
From the above analysis, neither [tex]\( f(x) \)[/tex] nor [tex]\( g(x) \)[/tex] are both negative at the same time in [tex]\((-\infty, -2) \)[/tex].
3. Consider the interval [tex]\((-2, 0)\)[/tex]:
For [tex]\( f(x) = x^2 + 2x \)[/tex]:
- Check the sign:
[tex]\[ x^2 + 2x = x(x + 2) \][/tex]
- For [tex]\( -2 < x < 0 \)[/tex], \
[tex]\( x \)[/tex] is negative and [tex]\( x + 2 \)[/tex] is positive. The product [tex]\( x(x + 2) \)[/tex] is negative.
For [tex]\( g(x) = \log(2x + 1) \)[/tex]:
- The argument must be positive:
[tex]\[ 2x + 1 > 0 \implies x > -\frac{1}{2} \][/tex]
- Thus, [tex]\((-2, -1/2) \subset (-2, 0)\)[/tex].
However since the domain [tex]\((-\infty, \infty)\)[/tex] is not valid in this interval.
Hence this is not the valid interval.
4. Consider the interval [tex]\((-1, 1)\)[/tex]:
For [tex]\( f(x) = x^2 + 2x \)[/tex]:
- The sign must be found within this interval:
[tex]\[ x^2 + 2x \ge 0 \implies x^2 + 2 \][/tex]
- For [tex]\( -1 < x < 1 \)[/tex],
[tex]\[ if x \in (-1, 0), x(x + 2) < -1 invalid domain \][/tex]
hence not a possible interval
5. Consider the interval [tex]\((-\infty, \infty)\)[/tex]:
For [tex]\( f(x) = x^2 + 2x \)[/tex]:
- Polynomial function across all values except needed in the possible onesbut,
invalid
In conclusion, the interval where both [tex]\( f(x) = x^2 + 2x \)[/tex] and [tex]\( g(x) = \log(2x + 1) \)[/tex] are negative is:
[tex]\[ (-\infty, -2) \][/tex]
1. Consider the interval [tex]\((-\infty, -2)\)[/tex]:
For [tex]\( f(x) = x^2 + 2x \)[/tex]:
- When [tex]\( x < -2 \)[/tex],
[tex]\( f(x) = x^2 + 2x \)[/tex] will be positive for [tex]\( x > -2 \)[/tex] and negative for [tex]\( x < -2 \)[/tex].
For [tex]\( g(x) = \log(2x + 1) \)[/tex]:
- The argument of the logarithm [tex]\( 2x + 1 \)[/tex] must be positive:
[tex]\[ 2x + 1 > 0 \implies x > -\frac{1}{2} \][/tex]
- However, since [tex]\( x < -2 \)[/tex], the argument [tex]\( 2x + 1 \)[/tex] will be less than [tex]\( 0 \)[/tex], making [tex]\( g(x) \)[/tex] undefined in this interval.
2. Consider the interval [tex]\((-\infty, -2)\)[/tex]:
For [tex]\( f(x) = x^2 + 2x \)[/tex]:
- When [tex]\( x < -2 \)[/tex],
[tex]\[ f(x) = x(x + 2) \][/tex]
- Since both [tex]\( x \)[/tex] and [tex]\( x+2 \)[/tex] are negative in the interval [tex]\((-\infty, -2) \)[/tex], [tex]\( f(x) \)[/tex] will be positive.
For [tex]\( g(x) = \log(2x + 1) \)[/tex]:
- The argument of [tex]\( g(x) \)[/tex] must be positive. For [tex]\( x < -2 \)[/tex], [tex]\( 2x + 1 < -3 \)[/tex] which makes the log function undefined or negative in the interval.
From the above analysis, neither [tex]\( f(x) \)[/tex] nor [tex]\( g(x) \)[/tex] are both negative at the same time in [tex]\((-\infty, -2) \)[/tex].
3. Consider the interval [tex]\((-2, 0)\)[/tex]:
For [tex]\( f(x) = x^2 + 2x \)[/tex]:
- Check the sign:
[tex]\[ x^2 + 2x = x(x + 2) \][/tex]
- For [tex]\( -2 < x < 0 \)[/tex], \
[tex]\( x \)[/tex] is negative and [tex]\( x + 2 \)[/tex] is positive. The product [tex]\( x(x + 2) \)[/tex] is negative.
For [tex]\( g(x) = \log(2x + 1) \)[/tex]:
- The argument must be positive:
[tex]\[ 2x + 1 > 0 \implies x > -\frac{1}{2} \][/tex]
- Thus, [tex]\((-2, -1/2) \subset (-2, 0)\)[/tex].
However since the domain [tex]\((-\infty, \infty)\)[/tex] is not valid in this interval.
Hence this is not the valid interval.
4. Consider the interval [tex]\((-1, 1)\)[/tex]:
For [tex]\( f(x) = x^2 + 2x \)[/tex]:
- The sign must be found within this interval:
[tex]\[ x^2 + 2x \ge 0 \implies x^2 + 2 \][/tex]
- For [tex]\( -1 < x < 1 \)[/tex],
[tex]\[ if x \in (-1, 0), x(x + 2) < -1 invalid domain \][/tex]
hence not a possible interval
5. Consider the interval [tex]\((-\infty, \infty)\)[/tex]:
For [tex]\( f(x) = x^2 + 2x \)[/tex]:
- Polynomial function across all values except needed in the possible onesbut,
invalid
In conclusion, the interval where both [tex]\( f(x) = x^2 + 2x \)[/tex] and [tex]\( g(x) = \log(2x + 1) \)[/tex] are negative is:
[tex]\[ (-\infty, -2) \][/tex]
