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Sagot :
To determine which function is increasing on the entire interval [tex]\((-\infty, \infty)\)[/tex], let's analyze each given function:
### Option A: [tex]\( h(x) = 2^x \)[/tex]
- This is an exponential function with base [tex]\(2\)[/tex], which is greater than [tex]\(1\)[/tex].
- Exponential functions with a base greater than [tex]\(1\)[/tex] are always increasing.
- Therefore, [tex]\( h(x) = 2^x \)[/tex] is increasing on the interval [tex]\((-\infty, \infty)\)[/tex].
### Option B: [tex]\( g(x) = -4(2^x) \)[/tex]
- This is also an exponential function with base [tex]\(2\)[/tex], but it is multiplied by [tex]\(-4\)[/tex].
- The multiplying factor [tex]\(-4\)[/tex] inverts the graph, making the function decreasing rather than increasing.
- Therefore, [tex]\( g(x) = -4(2^x) \)[/tex] is always decreasing.
### Option C: [tex]\( j(x) = x^2 + 8x + 1 \)[/tex]
- This is a quadratic function, and it can be rewritten in vertex form to analyze its behavior.
- Quadratic functions with positive leading coefficients (like this one, where the coefficient of [tex]\( x^2 \)[/tex] is [tex]\(1\)[/tex]) open upwards, meaning they have a minimum point and are not increasing over the entire interval.
- Specifically, they decrease on one side of the vertex and increase on the other side.
- Therefore, [tex]\( j(x) = x^2 + 8x + 1 \)[/tex] is not increasing on the entire interval.
### Option D: [tex]\( f(x) = -3x + 7 \)[/tex]
- This is a linear function with a slope of [tex]\(-3\)[/tex], which is negative.
- Linear functions with negative slopes decrease over the entire interval.
- Therefore, [tex]\( f(x) = -3x + 7 \)[/tex] is decreasing on the interval [tex]\((-\infty, \infty)\)[/tex].
### Conclusion:
From the analysis, the function that is increasing on the interval [tex]\((- \infty, \infty)\)[/tex] is:
- Option A: [tex]\( h(x) = 2^x \)[/tex]
Thus, the correct answer is:
[tex]\[ \boxed{1} \][/tex]
### Option A: [tex]\( h(x) = 2^x \)[/tex]
- This is an exponential function with base [tex]\(2\)[/tex], which is greater than [tex]\(1\)[/tex].
- Exponential functions with a base greater than [tex]\(1\)[/tex] are always increasing.
- Therefore, [tex]\( h(x) = 2^x \)[/tex] is increasing on the interval [tex]\((-\infty, \infty)\)[/tex].
### Option B: [tex]\( g(x) = -4(2^x) \)[/tex]
- This is also an exponential function with base [tex]\(2\)[/tex], but it is multiplied by [tex]\(-4\)[/tex].
- The multiplying factor [tex]\(-4\)[/tex] inverts the graph, making the function decreasing rather than increasing.
- Therefore, [tex]\( g(x) = -4(2^x) \)[/tex] is always decreasing.
### Option C: [tex]\( j(x) = x^2 + 8x + 1 \)[/tex]
- This is a quadratic function, and it can be rewritten in vertex form to analyze its behavior.
- Quadratic functions with positive leading coefficients (like this one, where the coefficient of [tex]\( x^2 \)[/tex] is [tex]\(1\)[/tex]) open upwards, meaning they have a minimum point and are not increasing over the entire interval.
- Specifically, they decrease on one side of the vertex and increase on the other side.
- Therefore, [tex]\( j(x) = x^2 + 8x + 1 \)[/tex] is not increasing on the entire interval.
### Option D: [tex]\( f(x) = -3x + 7 \)[/tex]
- This is a linear function with a slope of [tex]\(-3\)[/tex], which is negative.
- Linear functions with negative slopes decrease over the entire interval.
- Therefore, [tex]\( f(x) = -3x + 7 \)[/tex] is decreasing on the interval [tex]\((-\infty, \infty)\)[/tex].
### Conclusion:
From the analysis, the function that is increasing on the interval [tex]\((- \infty, \infty)\)[/tex] is:
- Option A: [tex]\( h(x) = 2^x \)[/tex]
Thus, the correct answer is:
[tex]\[ \boxed{1} \][/tex]
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