To determine which function is increasing on the interval [tex]\((-\infty, \infty)\)[/tex], we need to examine the behavior of the derivative of each function. A function is increasing on an interval if its derivative is positive over that entire interval. Let's analyze each function one by one.
### Option A: [tex]\( h(x) = 2^x - 1 \)[/tex]
- Derivative: The derivative of [tex]\( h(x) \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( h'(x) = \frac{d}{dx}(2^x - 1) \)[/tex].
- Calculation:
[tex]\[
h'(x) = \frac{d}{dx}(2^x) = 2^x \ln(2)
\][/tex]
- Behavior: Since [tex]\( 2^x \)[/tex] is always positive for all [tex]\( x \)[/tex] and [tex]\( \ln(2) \)[/tex] is a positive constant, [tex]\( h'(x) \)[/tex] is always positive.
### Option B: [tex]\( f(x) = -3x + 7 \)[/tex]
- Derivative: The derivative of [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( f'(x) = \frac{d}{dx}(-3x + 7) \)[/tex].
- Calculation:
[tex]\[
f'(x) = -3
\][/tex]
- Behavior: Since [tex]\( f'(x) = -3 \)[/tex] is constant and negative, the function [tex]\( f(x) \)[/tex] is always decreasing.
### Option C: [tex]\( j(x) = x^2 + 8x + 1 \)[/tex]
- Derivative: The derivative of [tex]\( j(x) \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( j'(x) = \frac{d}{dx}(x^2 + 8x + 1) \)[/tex].
- Calculation:
[tex]\[
j'(x) = 2x + 8
\][/tex]
- Behavior: The expression [tex]\( 2x + 8 \)[/tex] depends on [tex]\( x \)[/tex]. For [tex]\( x < -4 \)[/tex], [tex]\( j'(x) \)[/tex] is negative, and for [tex]\( x > -4 \)[/tex], [tex]\( j'(x) \)[/tex] is positive. Thus, [tex]\( j(x) \)[/tex] is not increasing over the entire interval [tex]\((-\infty, \infty)\)[/tex].
### Option D: [tex]\( g(x) = -4(2^x) \)[/tex]
- Derivative: The derivative of [tex]\( g(x) \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( g'(x) = \frac{d}{dx}(-4(2^x)) \)[/tex].
- Calculation:
[tex]\[
g'(x) = -4 \cdot \frac{d}{dx}(2^x) = -4 \cdot 2^x \ln(2)
\][/tex]
- Behavior: Since [tex]\( 2^x \)[/tex] is always positive for all [tex]\( x \)[/tex] and [tex]\( \ln(2) \)[/tex] is a positive constant, [tex]\( g'(x) = -4 \cdot 2^x \ln(2) \)[/tex] is always negative.
### Conclusion:
After analysing the derivatives of all the given functions, we find that:
- [tex]\( h'(x) = 2^x \ln(2) \)[/tex] is always positive.
- [tex]\( f'(x) = -3 \)[/tex] is always negative.
- [tex]\( j'(x) = 2x + 8 \)[/tex] changes sign depending on [tex]\( x \)[/tex].
- [tex]\( g'(x) = -4 \cdot 2^x \ln(2) \)[/tex] is always negative.
Therefore, the function that is increasing on the interval [tex]\((-\infty, \infty)\)[/tex] is:
Option A: [tex]\( h(x) = 2^x - 1 \)[/tex]
Correct Answer: A. [tex]\( h(x) = 2^x - 1 \)[/tex]
