To determine which function is increasing on the interval [tex]\( (-\infty, \infty) \)[/tex], we need to analyze each given function and check whether it consistently increases as [tex]\( x \)[/tex] increases.
Let's examine each function one by one:
### Function A: [tex]\( g(x) = -4 \cdot 2^x \)[/tex]
- Exponential functions of the form [tex]\( 2^x \)[/tex] are generally increasing, but since there is a negative coefficient multiplying the exponential term, this function will be decreasing. As [tex]\( x \)[/tex] increases, [tex]\( -4 \cdot 2^x \)[/tex] becomes more negative, meaning [tex]\( g(x) \)[/tex] decreases.
### Function B: [tex]\( f(x) = -3x + 7 \)[/tex]
- This is a linear function with a negative slope. In linear functions, the coefficient of [tex]\( x \)[/tex] determines the direction. Here, the slope is [tex]\(-3\)[/tex], which means the function decreases as [tex]\( x \)[/tex] increases.
### Function C: [tex]\( j(x) = x^2 + 8x + 1 \)[/tex]
- This is a quadratic function in the form [tex]\( ax^2 + bx + c \)[/tex] where [tex]\( a > 0 \)[/tex]. Quadratic functions open upwards and have a minimum point at [tex]\( x = -\frac{b}{2a} \)[/tex]. This function is not increasing everywhere; it decreases up to its vertex and then increases thereafter. Therefore, [tex]\( j(x) \)[/tex] is not increasing on the entire interval [tex]\( (-\infty, \infty) \)[/tex].
### Function D: [tex]\( h(x) = 2^x - 1 \)[/tex]
- Exponential functions of the form [tex]\( 2^x \)[/tex] are inherently increasing because [tex]\( 2 > 1 \)[/tex]. Adding or subtracting a constant does not change the increasing nature of the exponential term [tex]\( 2^x \)[/tex]. Therefore, [tex]\( 2^x - 1 \)[/tex] is increasing everywhere.
Based on the analysis of each function, the correct answer is:
[tex]\[
\boxed{h(x) = 2^x - 1}
\][/tex]
So, the function increasing on the interval [tex]\( (-\infty, \infty) \)[/tex] is [tex]\( h(x) = 2^x - 1 \)[/tex].
