Answer :
To determine which function is increasing on the interval [tex]\((-\infty, \infty)\)[/tex], let's analyze each one:
A. [tex]\( g(x) = -4(2^x) \)[/tex]
- The exponential function [tex]\(2^x\)[/tex] is increasing because as [tex]\(x\)[/tex] gets larger, [tex]\(2^x\)[/tex] also gets larger.
- However, multiplying by [tex]\(-4\)[/tex] changes the sign of the function, causing it to decrease as [tex]\(x\)[/tex] increases.
- Thus, [tex]\(g(x)\)[/tex] is not increasing on the interval [tex]\((-\infty, \infty)\)[/tex].
B. [tex]\( h(x) = 2^x - 1 \)[/tex]
- The exponential function [tex]\(2^x\)[/tex] is increasing because as [tex]\(x\)[/tex] gets larger, [tex]\(2^x\)[/tex] also gets larger.
- Subtracting a constant [tex]\(1\)[/tex] will shift the entire function down, but the overall trend of the function remains increasing.
- So, [tex]\(h(x)\)[/tex] is increasing on the interval [tex]\((-\infty, \infty)\)[/tex].
C. [tex]\( j(x) = x^2 + 8x + 1 \)[/tex]
- This is a quadratic function of the form [tex]\(j(x) = ax^2 + bx + c\)[/tex], where [tex]\(a = 1\)[/tex] (positive).
- Quadratic functions with a positive leading coefficient [tex]\(a\)[/tex] open upwards.
- They have a minimum point (vertex), causing them to decrease before the vertex and increase after the vertex.
- Thus, [tex]\(j(x)\)[/tex] is not increasing over the entire interval [tex]\((-\infty, \infty)\)[/tex].
D. [tex]\( f(x) = -3x + 7 \)[/tex]
- This is a linear function of the form [tex]\(f(x) = mx + b\)[/tex], where [tex]\(m = -3\)[/tex] (negative slope).
- A negative slope means the function decreases as [tex]\(x\)[/tex] increases.
- Therefore, [tex]\(f(x)\)[/tex] is not increasing on the interval [tex]\((-\infty, \infty)\)[/tex].
Therefore, the correct answer is:
B. [tex]\( h(x) = 2^x - 1 \)[/tex]
This is the only function among the given options that is increasing on the interval [tex]\((-\infty, \infty)\)[/tex].
A. [tex]\( g(x) = -4(2^x) \)[/tex]
- The exponential function [tex]\(2^x\)[/tex] is increasing because as [tex]\(x\)[/tex] gets larger, [tex]\(2^x\)[/tex] also gets larger.
- However, multiplying by [tex]\(-4\)[/tex] changes the sign of the function, causing it to decrease as [tex]\(x\)[/tex] increases.
- Thus, [tex]\(g(x)\)[/tex] is not increasing on the interval [tex]\((-\infty, \infty)\)[/tex].
B. [tex]\( h(x) = 2^x - 1 \)[/tex]
- The exponential function [tex]\(2^x\)[/tex] is increasing because as [tex]\(x\)[/tex] gets larger, [tex]\(2^x\)[/tex] also gets larger.
- Subtracting a constant [tex]\(1\)[/tex] will shift the entire function down, but the overall trend of the function remains increasing.
- So, [tex]\(h(x)\)[/tex] is increasing on the interval [tex]\((-\infty, \infty)\)[/tex].
C. [tex]\( j(x) = x^2 + 8x + 1 \)[/tex]
- This is a quadratic function of the form [tex]\(j(x) = ax^2 + bx + c\)[/tex], where [tex]\(a = 1\)[/tex] (positive).
- Quadratic functions with a positive leading coefficient [tex]\(a\)[/tex] open upwards.
- They have a minimum point (vertex), causing them to decrease before the vertex and increase after the vertex.
- Thus, [tex]\(j(x)\)[/tex] is not increasing over the entire interval [tex]\((-\infty, \infty)\)[/tex].
D. [tex]\( f(x) = -3x + 7 \)[/tex]
- This is a linear function of the form [tex]\(f(x) = mx + b\)[/tex], where [tex]\(m = -3\)[/tex] (negative slope).
- A negative slope means the function decreases as [tex]\(x\)[/tex] increases.
- Therefore, [tex]\(f(x)\)[/tex] is not increasing on the interval [tex]\((-\infty, \infty)\)[/tex].
Therefore, the correct answer is:
B. [tex]\( h(x) = 2^x - 1 \)[/tex]
This is the only function among the given options that is increasing on the interval [tex]\((-\infty, \infty)\)[/tex].
