Suppose that f(x) is a differentiable function and assume that g(x) is the inver

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). 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

Explanation / Answer

f(x) ==>slope f '(x)

f(x) ==>slope g '(x)

g(x) is the inverse function of f(x) ==>g(x)=f-1(x) ==>g'(x)=1/f '(f-1(x) )

L1(x) =f(a) +f '(a) (x-a)

L2(x) =g(b) +g'(b) (x-b)

b)g'(b)=1/f '(f-1(b) )

g'(b)=1/f '(a)

slopes are multiplicative inverses of each other

c)

L1(x) =f(a) +f '(a) (x-a) =b+f '(a) (x-a)

L2(x) =g(b) +g'(b) (x-b)=a+ g'(b) (x-b)=a+(1/f '(a))(x-b)

let L1(x) =b+f '(a) (x-a) =y ==>x=L-1(y)

==>L-1(y)=x=a+(y-b)/f '(a)

==>L-1(x)=a+(x-b)/f '(a)

==>L-1(x)=L2(x)

L2(x) is the inverse function of L1(x).

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