Problem 1. (1 point) Problem 4. (1 point) Algebraically find the inverse function of f (x) = 3-7et. For each of the given functions f(x), find the derivative (f-1) (c) at the given point c, first finding a = f (c). (See Theorem 7, f-1 ( x ) = page 156 of the Stewart Essential Calculus textbook) Graph f, f-and the line y = x on the same screen and check 1. f(x) = 6x+8x; c=-14 whether the graphs of f and f-1 are reflections about the line. a= ( f-1 )' ( c ) = Problem 2. (1 point) 2. f(x) =x2 - 16x+ 67 on the interval [8,co); c =7 Consider the function f (x) = vx -4. a = (A) Find f-1(4) = ( f- 1 )' ( c ) = B) Use Theorem 7, page 156 of the Stewart Essential Calcu- lus textbook to find (f-)'(4) (f -)' (4) = Problem 5. (1 point) (C) Calculate f-1(x) and state domain and range of f-1. Evaluate the following expressions. Use interval notation. If needed enter inf for co or -inf for -co. f-1(x) = (a) Ine' = Domain = Range = (b) eln3 = Calculate (f-!)'(4) from the formula for f-1(x) and check that it agrees with the result of part (B) (c) e2In(2) (d) In(1/e3) =. Problem 3. (1 point) Consider the function f (x) = - 8 for x > 1. (A) Find f-1 (2) = Problem 6. (1 point) Use the Laws of logarithms to rewrite the expression (B) Use Theorem 7, page 156 of the Stewart Essential Calcu- In(Vxy) lus textbook to find (f )'(2) ( f-1 )' (2 ) = in a form with no logarithm of a product, quotient or power. After rewriting we have (C) Calculate f-(x) and state domain and range of f-1. In(Vxy) = Aln(x) + BIn(y) Use interval notation. If needed enter inf for co or -inf for -co. f-1 ( x) = with the constant Domain = A= Range = and the constant Calculate (f-')'(2) from the formula for f (x) and check that it B =. agrees with the result of part (B)