To determine which function is increasing on the interval [tex]\((- \infty, \infty)\)[/tex], let's analyze each one step-by-step:
### Function A: [tex]\( f(x) = -3x + 7 \)[/tex]
This is a linear function with a slope of [tex]\( -3 \)[/tex]. The slope tells us the rate of change of the function:
- If the slope is positive, the function is increasing.
- If the slope is negative, the function is decreasing.
Since the slope here is [tex]\( -3 \)[/tex] (which is negative), [tex]\( f(x) \)[/tex] is a decreasing function over the interval [tex]\((- \infty, \infty)\)[/tex].
### Function B: [tex]\( j(x) = x^2 + 8x + 1 \)[/tex]
This is a quadratic function, which typically takes the form [tex]\( ax^2 + bx + c \)[/tex]:
- For quadratic functions, the sign of the coefficient [tex]\( a \)[/tex] determines whether the parabola opens upwards or downwards.
- Here, [tex]\( a = 1 \)[/tex] (positive), so the parabola opens upwards, which implies it has a minimum point.
To determine the interval where [tex]\( j(x) \)[/tex] is increasing, we can find the vertex (the minimum point):
[tex]\[ x = -\frac{b}{2a} = -\frac{8}{2 \cdot 1} = -4 \][/tex]
- For [tex]\( x < -4 \)[/tex], [tex]\( j(x) \)[/tex] is decreasing.
- For [tex]\( x > -4 \)[/tex], [tex]\( j(x) \)[/tex] is increasing.
However, [tex]\( j(x) \)[/tex] is not increasing over the entire interval [tex]\((- \infty, \infty) \)[/tex], only for [tex]\( x > -4 \)[/tex].
### Function C: [tex]\( g(x) = -4 \cdot 2^x \)[/tex]
This is an exponential function multiplied by a negative constant:
- The base of the exponential function [tex]\( 2^x \)[/tex] grows exponentially.
- Multiplying by a negative constant [tex]\( -4 \)[/tex] reverses the direction of growth.
As a result, [tex]\( g(x) \)[/tex] is a decreasing function over the interval [tex]\((- \infty, \infty)\)[/tex].
### Function D: [tex]\( h(x) = 2^x - 1 \)[/tex]
This is also an exponential function:
- The base [tex]\( 2^x \)[/tex] grows exponentially.
- Subtracting 1 does not affect the direction of the growth, only shifts the graph vertically.
Since the base [tex]\( 2 \)[/tex] is greater than 1, the function [tex]\( 2^x \)[/tex] is always increasing. Thus, [tex]\( h(x) = 2^x - 1 \)[/tex] is also increasing over the interval [tex]\((- \infty, \infty)\)[/tex].
### Conclusion
Among the given options, the function that is increasing on the interval [tex]\((- \infty, \infty)\)[/tex] is:
[tex]\[ \boxed{h(x) = 2^x - 1} \][/tex]
