Discover the best answers at Westonci.ca, where experts share their insights and knowledge with you. Explore thousands of questions and answers from a knowledgeable community of experts on our user-friendly platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
Sure, let's tackle this problem systematically by understanding both the original function and its inverse.
1. The original function given is:
[tex]\[ h(x) = -16x^2 + 25 \][/tex]
This function models the height [tex]\(h(x)\)[/tex] of a tennis ball [tex]\(x\)[/tex] seconds after it is dropped from a height of 25 feet.
2. To find the inverse function [tex]\(h^{-1}(x)\)[/tex], we need to express [tex]\(x\)[/tex] in terms of [tex]\(h\)[/tex]:
- Start by setting [tex]\(h(x)\)[/tex] equal to [tex]\(h\)[/tex]:
[tex]\[ h = -16x^2 + 25 \][/tex]
- Rearrange the equation to solve for [tex]\(x\)[/tex]:
[tex]\[ h - 25 = -16x^2 \][/tex]
[tex]\[ -16x^2 = h - 25 \][/tex]
[tex]\[ x^2 = \frac{25 - h}{16} \][/tex]
[tex]\[ x = \pm \sqrt{\frac{25 - h}{16}} \][/tex]
- Since time [tex]\(x\)[/tex] is non-negative, we take the positive root:
[tex]\[ x = \sqrt{\frac{25 - h}{16}} \][/tex]
3. Expressing this in terms of [tex]\(x\)[/tex], we can write the inverse function [tex]\(h^{-1}(x)\)[/tex]:
[tex]\[ h^{-1}(x) = \sqrt{\frac{25 - x}{16}} \][/tex]
4. The function [tex]\(h^{-1}(x)\)[/tex] is a square root function.
5. The domain of the inverse function [tex]\(h^{-1}(x)\)[/tex] must be within the range of the original function. Since [tex]\(h(x)\)[/tex] ranges from 0 to 25 (as it goes from the height of the roof down to the ground), [tex]\(x\)[/tex] must be within that range:
[tex]\[ 0 \leq x \leq 25 \][/tex]
To summarize:
- The inverse function will be a square root function and will have limits on the domain.
- Function [tex]\(h^{-1}\)[/tex] will give [tex]\(h^{-1}(x) = \sqrt{\frac{25 - x}{16}}\)[/tex] in terms of [tex]\(x\)[/tex].
Let's fill in the blanks based on this analysis:
Complete the statements about the inverse function, [tex]\(h^{-1}\)[/tex]:
- The inverse function will be a square root function and will have limits on the domain.
- Function [tex]\(h^{-1}\)[/tex] will give [tex]\(h^{-1}(x) = \sqrt{\frac{25 - x}{16}}\)[/tex] in terms of [tex]\(x\)[/tex].
1. The original function given is:
[tex]\[ h(x) = -16x^2 + 25 \][/tex]
This function models the height [tex]\(h(x)\)[/tex] of a tennis ball [tex]\(x\)[/tex] seconds after it is dropped from a height of 25 feet.
2. To find the inverse function [tex]\(h^{-1}(x)\)[/tex], we need to express [tex]\(x\)[/tex] in terms of [tex]\(h\)[/tex]:
- Start by setting [tex]\(h(x)\)[/tex] equal to [tex]\(h\)[/tex]:
[tex]\[ h = -16x^2 + 25 \][/tex]
- Rearrange the equation to solve for [tex]\(x\)[/tex]:
[tex]\[ h - 25 = -16x^2 \][/tex]
[tex]\[ -16x^2 = h - 25 \][/tex]
[tex]\[ x^2 = \frac{25 - h}{16} \][/tex]
[tex]\[ x = \pm \sqrt{\frac{25 - h}{16}} \][/tex]
- Since time [tex]\(x\)[/tex] is non-negative, we take the positive root:
[tex]\[ x = \sqrt{\frac{25 - h}{16}} \][/tex]
3. Expressing this in terms of [tex]\(x\)[/tex], we can write the inverse function [tex]\(h^{-1}(x)\)[/tex]:
[tex]\[ h^{-1}(x) = \sqrt{\frac{25 - x}{16}} \][/tex]
4. The function [tex]\(h^{-1}(x)\)[/tex] is a square root function.
5. The domain of the inverse function [tex]\(h^{-1}(x)\)[/tex] must be within the range of the original function. Since [tex]\(h(x)\)[/tex] ranges from 0 to 25 (as it goes from the height of the roof down to the ground), [tex]\(x\)[/tex] must be within that range:
[tex]\[ 0 \leq x \leq 25 \][/tex]
To summarize:
- The inverse function will be a square root function and will have limits on the domain.
- Function [tex]\(h^{-1}\)[/tex] will give [tex]\(h^{-1}(x) = \sqrt{\frac{25 - x}{16}}\)[/tex] in terms of [tex]\(x\)[/tex].
Let's fill in the blanks based on this analysis:
Complete the statements about the inverse function, [tex]\(h^{-1}\)[/tex]:
- The inverse function will be a square root function and will have limits on the domain.
- Function [tex]\(h^{-1}\)[/tex] will give [tex]\(h^{-1}(x) = \sqrt{\frac{25 - x}{16}}\)[/tex] in terms of [tex]\(x\)[/tex].
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
