Please help me solve these problems. 1.Find the inverse function (f^-1) of √3x-2
2. Integration
∫ the top is x while the bottom is 1. (sin(t))/t dt
#1 swap variables and solve for y:
x = √(3y-2)
x^2 = 3y-2
y = (x^2+2)/3
now f(x) has
domain: x >= 2/3
range: y >= 0
f^-1(x) has
domain: x >= 0
range: y >= 2/3
because f(x) has only one branch
#2
f(x) = ∫[1,x] sin(t)/t dt
I don't think you want f(x), because it is not elementary. But, using the 2nd fundamental theorem of calculus,
f'(x) = sin(x)/x