I'm confused on how to find the inverse function and solve?

(a) A way to evaluate #g(f(30)# is:

Start with:

#f(t) = 0.17t#

And evaluate it at #t = 30#:

#f(30) = 0.17(30)#

#f(30) = 5.1 " m"#

Then evaluate #g(r) = pir^2# at #r = 5.1 " m"#

#g(5.1" m") = pi(5.1" m")^2#

#g(5.1" m") = 26.01pi" m"^2#

(b) A way to find the inverse to a function (in an inverse exists) is:

Start with the function:

#f(t) = 0.17t#

Substitute #f^-1(r)# everywhere you see a t:

#f(f^-1(r)) = 0.17f^-1(r)#

Use a property that all inverses and their function must have #f(f^-1(r)) = r#:

#r = 0.17f^-1(r)#

Solve for #f^-1(r)#:

#f^-1(r) = r/0.17#

Now, evaluate #f^-1(r)# at #r = 4" m"#:

#f^-1(4" m") = (4" m")/0.17#

#f^-1(4) ~~ 25.2" s"#

I knew that the inverse function accepted, as its argument, a radius (in meters) and returned a value of time (in seconds), because the function accepts, as is argument, at time (in seconds) and returns a value of a radius (in meters). All inverses "undo" what the original function does.

(c) Find #g^-1(A)# using the same method that I used for #f^-1(r)#:

Start with #g(r)#

#g(r) = pir^2#

Substitute #g^-1(A)# everywhere you see and r:

#g(g^-1(A)) = pi(g^-1(A))^2#

Use the same property #g(g^-1(A)) = A#:

#A = pi(g^-1(A))^2#

Solve for #g^-1(A)#:

#A/pi = (g^-1(A))^2#

#(g^-1(A))^2= A/pi#

#g^-1(A)= +-sqrt(A/pi)#

All radii must be positive, therefore, we discard the #+-#

#g^-1(A)= sqrt(A/pi)#

Evaluate at #A = 100" m"^2#

#g^-1(100" m"^2) = sqrt((100" m"^2)/pi)#

#g^-1(100" m"^2) = 10/sqrt(pi)" m"#

I knew that the units using the same logic as in part (b).