Calculate g(b) and g'(b) where g is the inverse of f in the given domain. f(x) = √(x2 + 6x) for x = 0, b = 4.

My textbook offers nearly no help, and my notes are for different questions.
My textbook offers nearly no help, and my notes are for different questions.
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g(x) is the inverse of f(x)
y=f(x); g(y)=x; f(x)=√(x^{2} + 6x); x>=0
f'(x) = (1/2).(2.x + 6)/√(x^{2} + 6x)
b=4 => y=4
√(x^{2} + 6x) = 4
x^{2}+6x-16 = 0
x = (-6 ± √(36+4(16)))/2 => x = 2
=> g(4) = 2; (x,y)=(2,4)
d(g(y))/dx = d(x)/dx => g'(y).y'=1 => g'(f(x)) = 1/f'(x)
=> g'(4) = 1/f'(2) = 1/(5/4) = 4/5
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f(x) = √(x^{2}+ 6x)
Replace f(x) with y.
y = √(x^{2} + 6x)
Swap positions of x and y.
x = √(y^{2}+ 6y)
Square both sides.
x^{2} = y^{2}+6y
Add 9 to both sides.
x^{2}+9 = y^{2}+6y+9
Factorize right side.
x^{2}+9 = (y +3)^{2}
Take square root of both sides.
√(x^{2}+ 9) = y + 3
Subtract 3 from both sides.
y = -3 + √(x^{2}+ 9)
Replace y with g(x).
g(x) = -3 + √(x^{2} + 9)
Plug in value: x = 4
Simplify.
g(4) = -3 + √(25)
Simplify.
g(4) = -3 + 5
Simplify.
g(4) = 2
-----
Use properties of roots: √(a+b) = (a+b)^{1/2}
g(x) = -3 + (x^{2} + 9)^{1/2}
Take derivative of both sides.
[d/dx] [g(x)] = [d/dx] [-3 + (x^{2} + 9)^{1/2}]
Distribute [d/dx] into bracket (right side).
[d/dx] [g(x)] = [d/dx] [3] + (1/2) (x^{2}+ 9)^{-1/2} [d/dx] [x^{2} + 9]
Evaluate remaining derivatives.
g'(x) = 0 + (1/2) (x^{2} + 9)^{-1/2} (2x)
Simplify.
g'(x) = (x) (x^{2} + 9)^{-1/2}
Use properties of fractional exponents: (a+b)^{-1/2} = [1 / √(a+b)]
Plug in value: b = 4.
g'(4) = (4) / [√(4^{2} + 9)]
Simplify.
g'(4) = (4) / [√(25)]
Simplify.
g'(4) = 4/5
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