F⁻¹(x) stands for the invers of function f(x). The inverse of f(x) is equal to x. We need to solve for x.
solve for x
f(x) = 2x + 2
2x + 2 = f(x)
2x = f(x) - 2
x =
![\dfrac{f(x)-2}{2}](https://tex.z-dn.net/?f=%20%5Cdfrac%7Bf%28x%29-2%7D%7B2%7D)
f⁻¹(x) =
![\dfrac{x-2}{2}](https://tex.z-dn.net/?f=%20%5Cdfrac%7Bx-2%7D%7B2%7D)
Determine the value of f⁻¹(x) when x = 4
f⁻¹(x) =
![\dfrac{x-2}{2}](https://tex.z-dn.net/?f=%20%5Cdfrac%7Bx-2%7D%7B2%7D)
f⁻¹(4) =
![\dfrac{4-2}{2}](https://tex.z-dn.net/?f=%20%5Cdfrac%7B4-2%7D%7B2%7D)
f⁻¹(4) =
![\dfrac{2}{2}](https://tex.z-dn.net/?f=%20%5Cdfrac%7B2%7D%7B2%7D)
f⁻¹(4) = 1
When x = 4, the value of f⁻¹(x) is equal to 1
Answer:
The tenth place
Step-by-step explanation:
Answer:
The amount invested are $2600 and $11400 respectively
Step-by-step explanation:
<em>Let the first amount be x</em>
Given:
(First Investment)
Principal (P1) = x
Rate (R1) = 4.5%
Time (T) = 1 year
(Second Investment)
Principal (P2) = 4x + 1000
Rate (R2) = 6%
Time (T) = 1 year
Income = $801
Calculating the income from the first investment;
![I_1 = \frac{P_1R_1T}{100}](https://tex.z-dn.net/?f=I_1%20%3D%20%5Cfrac%7BP_1R_1T%7D%7B100%7D)
Substitute values for P1, R1 and T
![I_1 = \frac{x * 4.5 * 1}{100}](https://tex.z-dn.net/?f=I_1%20%3D%20%5Cfrac%7Bx%20%2A%204.5%20%2A%201%7D%7B100%7D)
![I_1 = \frac{4.5x}{100}](https://tex.z-dn.net/?f=I_1%20%3D%20%5Cfrac%7B4.5x%7D%7B100%7D)
Calculating the income from the second investment;
![I_2 = \frac{P_2R_2T}{100}](https://tex.z-dn.net/?f=I_2%20%3D%20%5Cfrac%7BP_2R_2T%7D%7B100%7D)
Substitute values for P2, R2 and T
![I_2 = \frac{(4x + 1000) * 6 * 1}{100}](https://tex.z-dn.net/?f=I_2%20%3D%20%5Cfrac%7B%284x%20%2B%201000%29%20%2A%206%20%2A%201%7D%7B100%7D)
![I_2 = \frac{6(4x + 1000)}{100}](https://tex.z-dn.net/?f=I_2%20%3D%20%5Cfrac%7B6%284x%20%2B%201000%29%7D%7B100%7D)
![I_1 + I_2 = Annual\ Income](https://tex.z-dn.net/?f=I_1%20%2B%20I_2%20%3D%20Annual%5C%20Income)
So:
![\frac{4.5x}{100} + \frac{6(4x + 1000)}{100} = 801](https://tex.z-dn.net/?f=%5Cfrac%7B4.5x%7D%7B100%7D%20%2B%20%5Cfrac%7B6%284x%20%2B%201000%29%7D%7B100%7D%20%3D%20801)
Multiply through by 100
![100 * \frac{4.5x}{100} +100 * \frac{6(4x + 1000)}{100} = 801 * 100](https://tex.z-dn.net/?f=100%20%2A%20%5Cfrac%7B4.5x%7D%7B100%7D%20%2B100%20%2A%20%20%5Cfrac%7B6%284x%20%2B%201000%29%7D%7B100%7D%20%3D%20801%20%2A%20100)
![4.5x +6(4x + 1000) = 801 * 100](https://tex.z-dn.net/?f=4.5x%20%2B6%284x%20%2B%201000%29%20%3D%20801%20%2A%20100)
![4.5x +24x + 6000 = 80100](https://tex.z-dn.net/?f=4.5x%20%2B24x%20%2B%206000%20%3D%2080100)
Collect Like Terms
![4.5x +24x = 80100 - 6000](https://tex.z-dn.net/?f=4.5x%20%2B24x%20%3D%2080100%20-%206000)
![28.5x = 74100](https://tex.z-dn.net/?f=28.5x%20%3D%2074100)
Divide through by 28.5
![x = \frac{74100}{28.5}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B74100%7D%7B28.5%7D)
![x = \$2600](https://tex.z-dn.net/?f=x%20%3D%20%5C%242600)
Recall that; the second invest
Amount Invested = 4x + 1000
This gives
![Amount = 4 * \$2600 + 1000](https://tex.z-dn.net/?f=Amount%20%3D%204%20%2A%20%5C%242600%20%2B%201000)
![Amount = \$11400](https://tex.z-dn.net/?f=Amount%20%3D%20%5C%2411400)
Hence;
<em>The amount invested are $2600 and $11400 respectively</em>
$7680:100%=$x:120.45%, 100x=7680*120.45, x=(7680*120.45)/100, x=$9250.56
Cost of the pool with interest rate is $9250.56.
3yr*12mo=36mo
$9250.56/36mo=256.96$/mo
<span> Charlotte's monthly payment will be $256.96.</span>