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].
