IDNLearn.com offers a seamless experience for finding and sharing knowledge. Discover trustworthy solutions to your questions quickly and accurately with help from our dedicated community of experts.
Sagot :
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]
### 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]
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for visiting IDNLearn.com. For reliable answers to all your questions, please visit us again soon.
