IDNLearn.com offers a unique blend of expert answers and community-driven knowledge. Whether it's a simple query or a complex problem, our community has the answers you need.
Sagot :
Let's analyze the given functions [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex] and [tex]\( g(x) = 2^x \)[/tex] to identify their key features step by step.
### 1. Analyze Decrease Over the Interval [tex]\([1, \infty)\)[/tex]:
- For [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- The function is a downward-opening parabola (since the coefficient of [tex]\( x^2 \)[/tex] is negative).
- To confirm decreasing behavior, complete the square or find the vertex:
[tex]\[ f(x) = -(x^2 - 2x - 1) = -(x^2 - 2x + 1 - 1) = -(x - 1)^2 + 2. \][/tex]
- The vertex is at [tex]\(x = 1\)[/tex] and the maximum value is at [tex]\( f(1) = 2 \)[/tex]. Therefore, [tex]\( f(x) \)[/tex] is decreasing for [tex]\( x \geq 1 \)[/tex].
- For [tex]\( g(x) = 2^x \)[/tex]:
- [tex]\( g(x) \)[/tex] is an exponential function with base 2.
- The function [tex]\( g(x) \)[/tex] is increasing for all [tex]\( x \)[/tex], so it doesn't decrease for any interval.
Thus, the claim that both functions decrease over the interval [tex]\([1, \infty)\)[/tex] is false.
### 2. Compare the Range:
- Range of [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- The maximum value occurs at [tex]\( x = 1 \)[/tex], [tex]\( f(1) = 2 \)[/tex].
- Since it opens downwards, the range of [tex]\( f(x) \)[/tex] is:
[tex]\[ (-\infty, 2]. \][/tex]
- Range of [tex]\( g(x) = 2^x \)[/tex]:
- Since [tex]\( g(x) \)[/tex] is an exponential function, the range is:
[tex]\[ (0, \infty). \][/tex]
Thus, the claim that both functions have the same range of [tex]\((- \infty, 2)\)[/tex] is false.
### 3. Compare the [tex]\( x \)[/tex]-Intercepts:
- [tex]\( x \)[/tex]-Intercept of [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ -x^2 + 2x + 1 = 0. \][/tex]
- Solving the quadratic equation:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{4 + 4}}{-2}. \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{8}}{-2} = 1 \pm \sqrt{2}. \][/tex]
- The [tex]\( x \)[/tex]-intercepts are [tex]\( x = 1 + \sqrt{2} \)[/tex] and [tex]\( x = 1 - \sqrt{2} \)[/tex].
- [tex]\( x \)[/tex]-Intercept of [tex]\( g(x) = 2^x \)[/tex]:
- Set [tex]\( g(x) = 0 \)[/tex]:
- The exponential function [tex]\( 2^x \)[/tex] never crosses the x-axis (it never equals zero), so there is no [tex]\( x \)[/tex]-intercept.
Thus, the claim that both functions have the same [tex]\( x \)[/tex]-intercept of [tex]\((-1, 0)\)[/tex] is false.
### 4. Compare the [tex]\( y \)[/tex]-Intercepts:
- [tex]\( y \)[/tex]-Intercept of [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- Set [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -0^2 + 2 \cdot 0 + 1 = 1. \][/tex]
- The [tex]\( y \)[/tex]-intercept is [tex]\( (0, 1) \)[/tex].
- [tex]\( y \)[/tex]-Intercept of [tex]\( g(x) = 2^x \)[/tex]:
- Set [tex]\( x = 0 \)[/tex]:
[tex]\[ g(0) = 2^0 = 1. \][/tex]
- The [tex]\( y \)[/tex]-intercept is [tex]\( (0, 1) \)[/tex].
Thus, the claim that both functions have the same [tex]\( y \)[/tex]-intercept of [tex]\((0, 1)\)[/tex] is true.
### Conclusion:
The only correct statement is:
- Both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] have the same [tex]\( y \)[/tex]-intercept of [tex]\((0, 1)\)[/tex].
### 1. Analyze Decrease Over the Interval [tex]\([1, \infty)\)[/tex]:
- For [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- The function is a downward-opening parabola (since the coefficient of [tex]\( x^2 \)[/tex] is negative).
- To confirm decreasing behavior, complete the square or find the vertex:
[tex]\[ f(x) = -(x^2 - 2x - 1) = -(x^2 - 2x + 1 - 1) = -(x - 1)^2 + 2. \][/tex]
- The vertex is at [tex]\(x = 1\)[/tex] and the maximum value is at [tex]\( f(1) = 2 \)[/tex]. Therefore, [tex]\( f(x) \)[/tex] is decreasing for [tex]\( x \geq 1 \)[/tex].
- For [tex]\( g(x) = 2^x \)[/tex]:
- [tex]\( g(x) \)[/tex] is an exponential function with base 2.
- The function [tex]\( g(x) \)[/tex] is increasing for all [tex]\( x \)[/tex], so it doesn't decrease for any interval.
Thus, the claim that both functions decrease over the interval [tex]\([1, \infty)\)[/tex] is false.
### 2. Compare the Range:
- Range of [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- The maximum value occurs at [tex]\( x = 1 \)[/tex], [tex]\( f(1) = 2 \)[/tex].
- Since it opens downwards, the range of [tex]\( f(x) \)[/tex] is:
[tex]\[ (-\infty, 2]. \][/tex]
- Range of [tex]\( g(x) = 2^x \)[/tex]:
- Since [tex]\( g(x) \)[/tex] is an exponential function, the range is:
[tex]\[ (0, \infty). \][/tex]
Thus, the claim that both functions have the same range of [tex]\((- \infty, 2)\)[/tex] is false.
### 3. Compare the [tex]\( x \)[/tex]-Intercepts:
- [tex]\( x \)[/tex]-Intercept of [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ -x^2 + 2x + 1 = 0. \][/tex]
- Solving the quadratic equation:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{4 + 4}}{-2}. \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{8}}{-2} = 1 \pm \sqrt{2}. \][/tex]
- The [tex]\( x \)[/tex]-intercepts are [tex]\( x = 1 + \sqrt{2} \)[/tex] and [tex]\( x = 1 - \sqrt{2} \)[/tex].
- [tex]\( x \)[/tex]-Intercept of [tex]\( g(x) = 2^x \)[/tex]:
- Set [tex]\( g(x) = 0 \)[/tex]:
- The exponential function [tex]\( 2^x \)[/tex] never crosses the x-axis (it never equals zero), so there is no [tex]\( x \)[/tex]-intercept.
Thus, the claim that both functions have the same [tex]\( x \)[/tex]-intercept of [tex]\((-1, 0)\)[/tex] is false.
### 4. Compare the [tex]\( y \)[/tex]-Intercepts:
- [tex]\( y \)[/tex]-Intercept of [tex]\( f(x) = -x^2 + 2x + 1 \)[/tex]:
- Set [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -0^2 + 2 \cdot 0 + 1 = 1. \][/tex]
- The [tex]\( y \)[/tex]-intercept is [tex]\( (0, 1) \)[/tex].
- [tex]\( y \)[/tex]-Intercept of [tex]\( g(x) = 2^x \)[/tex]:
- Set [tex]\( x = 0 \)[/tex]:
[tex]\[ g(0) = 2^0 = 1. \][/tex]
- The [tex]\( y \)[/tex]-intercept is [tex]\( (0, 1) \)[/tex].
Thus, the claim that both functions have the same [tex]\( y \)[/tex]-intercept of [tex]\((0, 1)\)[/tex] is true.
### Conclusion:
The only correct statement is:
- Both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] have the same [tex]\( y \)[/tex]-intercept of [tex]\((0, 1)\)[/tex].
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. For precise answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.
