Question

5. Let f(x) = arctan(In x) for all x >0. A graph of y = f(x) is shown in the figure. (a) Find the formula for the derivative f'(x). Then explain how you can deduce from this formula that f is invertible. (b) Find the formula for f-1(x), the inverse of f. (c) What is the domain and range of f-1? (d) Sketch a graph of the function y=f-1(x). (e) Now determine the value of (F-1)(0) using your results from (a) and (b) and the Derivative Theorem: 1 (F-1)(0) = f'(f-(0) KI y = f(x) -2

Answer 1

f(x)= arctan (hux) x 70 = tant (mx), x70 @ f'(x)=d tant ( mx) 1 (mm) U da It (mx)2 Chlmx)=jú 1 x{1+[mx)"} then we If we assume that f' is not invertible can conclude that it must not be One-one, because the range and codomain can be easily taken care of when it comes to invertible for, we can choose range= codomain for an One-One function and make it inventible. We shall assume that So, here in our case if' is not one-one. So, 7 x X₂ such that f(x) = f (22) And now apply Rolle's theorem to the function 'f' on the interval [x].

. f is continuous on [x1, 2] f is differentiable on (12) f(x1) = f (22) So, there exists atleast one point o such that > f'(c)=0 ow C (1+(me) and this is not possible C70. "f' must be one-one & hence of' is invertible. b To determine f (x). y = tant (mx) ye (-12, 17) tanya ha Ina e tany- e x = & tany tany f(y) = ye(-">72) Answer taux Domain of f(x) = e is (-1/2, 1/2) and the range of f(x) = 0 tanx is (0,x)

tana The graph of e g(x), HE(-712, 7/2). y=etanx ده) -12 1/2 tan(o) re f'(o)= et = eo =1 4 f'() - 2 (1 + (0)2 = 1 1 (f) (0) f' (f-(0) 2 f'(1) -1 Answer