a) To find (f^−1)'(a), where f(x) = 5 + x^2 + tan(πx/2) and a = 5, we need to find the derivative of the inverse function at x = a.
Since a = 5, we want to find (f^−1)'(5).
To find the inverse function, we can switch x and y in the original equation:
x = 5 + y^2 + tan(πy/2).
Now we solve for y:
x - 5 = y^2 + tan(πy/2).
The derivative of the inverse function is given by (f^−1)'(5) = 1 / (f'(f^−1(5)).
To find f^−1(5), we substitute x = 5 into the equation x = 5 + y^2 + tan(πy/2):
5 = 5 + y^2 + tan(πy/2).
Simplifying, we get y^2 + tan(πy/2) = 0.
Next, we find the derivative of f(x) = 5 + x^2 + tan(πx/2):
f'(x) = 2x + πsec^2(πx/2).
Substituting f^−1(5) into f'(x), we get:
f'(f^−1(5)) = 2f^−1(5) + πsec^2(πf^−1(5)/2).
Finally, we can calculate (f^−1)'(5):
(f^−1)'(5) = 1 / (f'(f^−1(5))).
b) To find a formula for the inverse of f(x) = 2x - 1 / (2x + 3), we switch x and y in the equation and solve for y:
x = 2y - 1 / (2y + 3).
Multiply both sides by (2y + 3) to get rid of the fraction:
x(2y + 3) = 2y - 1.
Expand and rearrange the equation:
2xy + 3x = 2y - 1.
Move all terms involving y to one side:
2xy - 2y = -3x - 1.
Factor out y:
y(2x - 2) = -3x - 1.
Divide both sides by (2x - 2):
y = (-3x - 1) / (2x - 2).
Thus, the inverse of f(x) = 2x - 1 / (2x + 3) is f^−1(x) = (-3x - 1) / (2x - 2).
c) To find a formula for the inverse of f(x) = 2 + sqrt(4 + 7x), we switch x and y in the equation and solve for y:
x = 2 + sqrt(4 + 7y).
Move 2 to the other side:
x - 2 = sqrt(4 + 7y).
Square both sides:
(x - 2)^2 = 4 + 7y.
Expand and rearrange the equation:
x^2 - 4x + 4 = 4 + 7y.
Move all terms involving y to one side:
7y = x^2 - 4x.
Divide both sides by 7:
y = (x^2 - 4x) / 7.
Thus, the inverse of f(x) = 2 + sqrt(4 + 7x) is f^−1(x) = (x^2 - 4x) / 7.