Theorem 4.4.3. Suppose that f is continuous and one-to-one on[a,b]. Then f1 is continuous on f([a,b]).Proof. From the previous theorem, f is either strictly increasing ordecreasing on [a, b]. Assume that f is strictly increasing. (Applyingthese considerations to the function -f if f were decreasing wouldcomplete the proof.) Take any point yo E (f(a), f(b)) and any ɛ > 0.We need to show that there is a ō> 0 such thatly– yol < 8implies w)-f0)| < e.Set xo = f'(yo) E (a,b). Replacing the given ɛ with a smaller valueif necessary, we can assume that a s xo - E < Xo + ɛ < b. Then f(xo- E) < f(xo) < f(Xo + ɛ), and from the previous theorem, f1 is strictlyincreasing on [f(Xo – €), f(xo + ɛ)], mapping this interval into [Xo – E,Xo + ɛ]. Hence, let|8 = Min {yo - f(xo – €), f(xo+E) – yo}(see Figure 4.14). If yo =argument to the interval [a, a + ɛ] (or [b – ɛ,b]).f(a) (or f(b)), we modify the aboveThus, we have that, for a function f one-to-one on an interval [a,b], continuity of f implies the continuity of the inverse. We now usethis result in our discussion on differentiability. 2. Define, for x e S = [0,1] U [2,3), the function%3D0

Given, f=x0≤x≤14-x2≤x<3 S=[0,1] ∪ [2,3) (a). Since the function is linear on S, the function is…

Answered: 2. Define, for x E S [0,1] U [2,3), the…

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2. Define, for x E S [0,1] U [2,3), the 0

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Theorem 4.4.3. Suppose that f is continuous and one-to-one on
[a,b]. Then f1 is continuous on f([a,b]).
Proof. From the previous theorem, f is either strictly increasing or
decreasing on [a, b]. Assume that f is strictly increasing. (Applying
these considerations to the function -f if f were decreasing would
complete the proof.) Take any point yo E (f(a), f(b)) and any ɛ > 0.
We need to show that there is a ō> 0 such that
ly– yol < 8
implies w)-f0)| < e.
Set xo = f'(yo) E (a,b). Replacing the given ɛ with a smaller value
if necessary, we can assume that a s xo - E < Xo + ɛ < b. Then f(xo
- E) < f(xo) < f(Xo + ɛ), and from the previous theorem, f1 is strictly
increasing on [f(Xo – €), f(xo + ɛ)], mapping this interval into [Xo – E,
Xo + ɛ]. Hence, let
|
8 = Min {yo - f(xo – €), f(xo+E) – yo}
(see Figure 4.14). If yo =
argument to the interval [a, a + ɛ] (or [b – ɛ,b]).
f(a) (or f(b)), we modify the above
Thus, we have that, for a function f one-to-one on an interval [a,
b], continuity of f implies the continuity of the inverse. We now use
this result in our discussion on differentiability.

2. Define, for x e S = [0,1] U [2,3), the function
%3D
0<x<1
f(x) =
14-x,
2<x<3'
(a) Show that f is continuous and one-to-one on S.
(b) Show that T = f(S) = [0,2],
(c) Determine f'(y) for y e T.
(d) Show that f1 is not continuous at 1.
(e) Why doesn't this example contradict Theorem 4.4.3?

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