In Problems 21-28 , determine whether f ( t ) is continuous, piecewise continuous, or neither on [ 0 , 10 ] and sketch the graph of f ( t ) . f ( t ) = { 1 / t , 0 < t < 1 , 1 , 1 ≤ t ≤ 2 , 1 − t , 2 < t ≤ 10
In Problems 21-28 , determine whether f ( t ) is continuous, piecewise continuous, or neither on [ 0 , 10 ] and sketch the graph of f ( t ) . f ( t ) = { 1 / t , 0 < t < 1 , 1 , 1 ≤ t ≤ 2 , 1 − t , 2 < t ≤ 10
Solution Summary: The author explains that the function f(t) is neither continuous nor piecewise continuous on any interval.
B. The graph of f(x) is given by
(-14)
(1,4)
(-1,–2)
(5,-2)
(3, -4)
(4, 5)
Determine is f is continuous at x = c.
(a) x = -4
(c) x = 1
(e) x = 5
(b) x = -1
(d) z 3
2. Let f be a continuous function defined on a closed, bounded interval I = [a, 6].
Assume that f is one-to-one. Let m (M, respectively) be the minimum (maximum,
respectively) of f. Then by problem 1, we know that either f(a) = m and f(b) = M;
or f(a) = M and f(b) = m. If f(a)
monotone increasing. If f(a) = M and f(a)
monotone decreasing.
%3D
%3D
= m and f(b) = M, then show that f is strictly
= m, then show that f is strictly
%3D
Hint: you need to use the Intermediate Value Theorem and argue by contradition.
Remark: Problem 2 basically says the only way for a continuous func-
tion to be one-to-one is to be strictly monotone. In other words, only
strictly monotone continuous functions can have inverse. This is actually
a Theorem in our text.
4. Let function ƒ: R → R and function g: R → R be two functions for which f(g(x)) = x and g(f(x)) = x. Through some combination of pictures
and/or words, explain the following: if f(a) is equal to b and ƒ' (a) is equal to m (where m is nonzero), then it follows that g′ (b) is equal to
Chapter 7 Solutions
Fundamentals of Differential Equations and Boundary Value Problems
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