Suppose that f(x) is a differentiable function and assume that g(x) is the inverse function of f(x).?

Question

Suppose that f(x) is a differentiable function and assume that g(x) is the inverse
function of f(x). Let L1(x) be the linearization of f(x) at x = a and let L2(x) be the
linearization of g(x) at x = b where b = f(a).
In the following problems, use the results of Problem 1, the formula for linearization in Section
4.1, and the formula for the derivative of the inverse of a function in Section 3.8 of the
textbook. Figure 1 of Section 3.8 might help you to visualize what is going on geometrically.
(a) Write the formulas for L1(x) and L2(x).

(b) How are the slopes of L1(x) and L2(x) related?

(c) If the graph of L1(x) is not a horizontal line, then show that L2(x) is the inverse function of L1(x). (It is helpful to use that b = f(a) and a = g(b) in this problem.)

(d) If f'(a) doesn't equal ±1, explain why the graphs of L1(x) and L2(x) intersect on the line given by y = x

Answer 1

"linearization in Section 4.1"

" Figure 1 of Section 3.8 "

in your book

Answer 2

　

b = f(a)

a = g(b)



a)



L₁(x) = f(a) + f'(a) (x − a)

L₁(x) = b + f'(a) (x − a)



L₂(x) = g(b) + g'(b) (x − b)

L₂(x) = a + g'(b) (x − b)



b)



The slopes are reciprocals of each other: g'(b) = 1/f'(a)



c)



L₁ and L₂ are inverse functions if L₁(L₂(x)) = L₂(L₁(x)) = x



L₁(L₂(x))

= b + f'(a) (L₂(x) − a)

= b + f'(a) (a + g'(b) (x − b) − a)

= b + f'(a) * g'(b) (x − b)

= b + f'(a) * 1/f'(a) (x − b)

= b + (x − b)

= x



L₂(L₁(x))

= a + g'(b) (L₁(x) − b)

= a + g'(b) (b + f'(a) (x − a) − b)

= a + g'(b) * f'(a) (x − a)

= a + 1/f'(a) * f'(a) (x − a)

= a + (x − a)

= x



d)



L₁(x) = L₂(x)

b + f'(a) (x − a) = a + g'(b) (x − b)

b + f'(a) (x − a) = a + 1/f'(a) (x − b)

b f'(a) + f'(a)² (x − a) = a f'(a) + (x − b)

b f'(a) + x f'(a)² − a f'(a)² = a f'(a) + x − b

x (f'(a)² − 1) = a f'(a)² + a f'(a) − b f'(a) − b

x (f'(a) − 1) (f'(a) + 1) = a f'(a) (f'(a) + 1) − b (f'(a) + 1)

x (f'(a) − 1) (f'(a) + 1) = (f'(a) + 1) (a f'(a) − b)



Since f'(a) ≠ ±1, then divide both sides by (f'(a) − 1) (f'(a) + 1)



x = (a f'(a) − b) / (f'(a) − 1)



y = L₁(x)

y = b + f'(a) (x − a)

y = b + f'(a) ((a f'(a) − b) / (f'(a) − 1) − a)

y = b + f'(a) (a f'(a) − b − a f'(a) + a) / (f'(a) − 1)

y = b + f'(a) (a − b) / (f'(a) − 1)

y = (b f'(a) − b + a f'(a) − b f'(a)) / (f'(a) − 1)

y = (a f'(a) − b) / (f'(a) − 1)

y = x



P.S. It's no use including information about: Problem 1, Section 4.1, Section 3.8, Figure 1.



Do you suppose we all have a copy of your textbook, or that we even know what textbook you are using?

Answer 3

Since we haven't got access to your textbook, it is ridiculous that you would expect us to "use the results of Problem 1, the formula...in Section 4.1, and the formula...in Section 3.8" etc. Not to mention "Figure 1 of Section 3.8"