Find the inverses of each of the functions below algebraically.
p(r)=2r^2+2r−1
3y+5x=18
can you find the other one aswell please
p(r) or p = 2r2 + 2r - 1
p = 2(r+0.5)2 - 1.5
r = 2(p+0.5)2 - 1.5
(r+1.5)/2 = (p+0.5)2
+-sqrt[(r+1.5)/2] = p + 0.5
p-1(r) = -0.5 +- sqrt[(r+1.5)/2]
After testing, only right-hand side works. Hence p-1(r) = -0.5 + sqrt[(r+1.5)/2]
Virtual high school right?
To find the inverse of a function algebraically, you need to follow these steps:
1. Replace the function notation with the variable 'y.' For the first equation, replace p(r) with y: y = 2r^2 + 2r - 1.
For the second equation, rearrange it to solve for y: 3y + 5x = 18.
2. Swap the variables 'x' and 'y.' This step is done to interchange the roles of the dependent and independent variables.
For the first equation, replace 'y' with 'r' and 'r' with 'y': r = 2y^2 + 2y - 1.
For the second equation, interchange 'x' and 'y': 3x + 5y = 18.
3. Solve the new equation obtained for 'y.' This step involves rearranging the equation to solve for 'y.'
Let's solve each equation step by step:
1. For the first equation, y = 2r^2 + 2r - 1.
2. Swap the variables: r = 2y^2 + 2y - 1.
3. Solve for 'y':
To solve, bring all terms to one side of the equation: 2y^2 + 2y - 1 - r = 0.
Use the quadratic formula: y = (-b ± √(b^2 - 4ac)) / 2a.
In this case, a = 2, b = 2, and c = -r.
Substituting these values into the formula, we get:
y = (-2 ± √(2^2 - 4 * 2 * (-r))) / (2 * 2).
Simplifying further, y = (-2 ± √(4 + 16r)) / 4.
Splitting into two cases:
Case 1: y = (-2 + √(4 + 16r)) / 4.
Case 2: y = (-2 - √(4 + 16r)) / 4.
Thus, the inverse function of p(r) = 2r^2 + 2r - 1 is:
f(r) = (-2 + √(4 + 16r)) / 4 or f(r) = (-2 - √(4 + 16r)) / 4.
Now, let's work on the second equation:
1. For the second equation, 3y + 5x = 18.
2. Swap the variables: 3x + 5y = 18.
3. Solve for 'y':
To solve, bring all terms to one side of the equation: 5y = 18 - 3x.
Divide both sides by 5: y = (18 - 3x) / 5.
Thus, the inverse function of 3y + 5x = 18 is:
f(x) = (18 - 3x) / 5.