Given the function for velocity [tex]\( V \)[/tex] (in centimeters per second) of blood in an artery at a distance [tex]\( x \)[/tex] (in centimeters) from the center of the artery:
[tex]\[ V = f(x) = 500 \left(0.04 - x^2\right) \][/tex]
We need to find the inverse function, domain for the inverse function, interpret the inverse function, and calculate the distance based on a given velocity.
### Part a: Find the inverse function [tex]\( f^{-1}(V) \)[/tex]
1. Start with the given function:
[tex]\[ V = 500 \left(0.04 - x^2\right) \][/tex]
2. Simplify and solve for [tex]\( x \)[/tex] in terms of [tex]\( V \)[/tex]:
[tex]\[ V = 500 \left(0.04 - x^2\right) \][/tex]
3. Divide by 500 to isolate the quadratic term:
[tex]\[ \frac{V}{500} = 0.04 - x^2 \][/tex]
4. Move [tex]\( x^2 \)[/tex] to the other side:
[tex]\[ x^2 = 0.04 - \frac{V}{500} \][/tex]
5. Take the square root of both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \pm\sqrt{0.04 - \frac{V}{500}} \][/tex]
Given that [tex]\( 0 \leq x \leq 0.2 \)[/tex] in the initial domain, we take the positive root:
[tex]\[ x = \sqrt{0.04 - \frac{V}{500}} \][/tex]
So, the inverse function is:
[tex]\[ f^{-1}(V) = \sqrt{0.04 - \frac{V}{500}} \][/tex]
### Part b: Write the domain for the inverse function in interval notation.
To find the domain for the inverse function [tex]\( f^{-1} \)[/tex], we need to consider the range of the original function [tex]\( f(x) \)[/tex]. Let's explore:
1. At [tex]\( x = 0 \)[/tex]:
[tex]\[ V = 500 \left(0.04 - 0^2\right) = 500(0.04) = 20 \][/tex]
2. At [tex]\( x = 0.2 \)[/tex]:
[tex]\[ V = 500 \left(0.04 - (0.2)^2\right) = 500 \left(0.04 - 0.04\right) = 500(0) = 0 \][/tex]
Thus, the range of [tex]\( V \)[/tex] is [tex]\([0, 20]\)[/tex], which becomes the domain for [tex]\( f^{-1}(V) \)[/tex].
So, the domain for the inverse function in interval notation is:
[tex]\[ [0, 20] \][/tex]
### Part c: Interpret what the inverse function is used for.
The inverse function determines the distance from the center of the artery given the velocity of the blood.
### Part d: Find the distance from the center of an artery with a velocity of [tex]\( 8 \text{ cm/s} \)[/tex].
Using the inverse function [tex]\( f^{-1}(V) \)[/tex], substitute [tex]\( V = 8 \)[/tex]:
[tex]\[ x = \sqrt{0.04 - \frac{8}{500}} \][/tex]
[tex]\[ x = \sqrt{0.04 - 0.016} \][/tex]
[tex]\[ x = \sqrt{0.024} \][/tex]
Now, calculate the square root and round to 2 decimal places:
[tex]\[ x \approx \sqrt{0.024} \approx 0.155 \][/tex]
Therefore, the distance from the center of an artery with a velocity of [tex]\( 8 \text{ cm/s} \)[/tex] is approximately:
[tex]\[ 0.15 \text{ cm} \][/tex]
