1. Prove that if f is continuous on [a, b) and lim f(x) as x goes to -infinity e

Question

1. Prove that if f is continuous on [a, b) and lim f(x) as x goes to -infinity exists, then f is bounded and uniformly continuous on [a, b).
2. Let f: [0, 2] going to real numbers be a continuous function. show that a) if f(0)=f(2), then there is x1, x2 in [0, 2] such that x2-x1=1 and f(x1)=f(x2). (hint use g(x)=f(x+1)-f(x0 on [0, 1] ) b)if f ( [0, 2] ) subset of [0, 2], then there is c in [0, 2] with f(c) =c. 3. Suppose f is continuous on [a, b]. show that the functions defined by m(x)=inf{ f(y): y in [a, x]} and M(x)=sup{ f(y): y in [a, x]} are defined well and are also continuous on [a, b]. 1. Prove that if f is continuous on [a, b) and lim f(x) as x goes to -infinity exists, then f is bounded and uniformly continuous on [a, b).
2. Let f: [0, 2] going to real numbers be a continuous function. show that a) if f(0)=f(2), then there is x1, x2 in [0, 2] such that x2-x1=1 and f(x1)=f(x2). (hint use g(x)=f(x+1)-f(x0 on [0, 1] ) b)if f ( [0, 2] ) subset of [0, 2], then there is c in [0, 2] with f(c) =c. 3. Suppose f is continuous on [a, b]. show that the functions defined by m(x)=inf{ f(y): y in [a, x]} and M(x)=sup{ f(y): y in [a, x]} are defined well and are also continuous on [a, b]. 1. Prove that if f is continuous on [a, b) and lim f(x) as x goes to -infinity exists, then f is bounded and uniformly continuous on [a, b).
2. Let f: [0, 2] going to real numbers be a continuous function. show that a) if f(0)=f(2), then there is x1, x2 in [0, 2] such that x2-x1=1 and f(x1)=f(x2). (hint use g(x)=f(x+1)-f(x0 on [0, 1] ) b)if f ( [0, 2] ) subset of [0, 2], then there is c in [0, 2] with f(c) =c. 3. Suppose f is continuous on [a, b]. show that the functions defined by m(x)=inf{ f(y): y in [a, x]} and M(x)=sup{ f(y): y in [a, x]} are defined well and are also continuous on [a, b].

Explanation / Answer

2) (a)let g(x) = f(x+1) - f(x)

g(0) = f(1) - f(0)

g(1) = f(2) - f(1)

But f(0) = f(2)

So g(0) = - g(1)

i.e g(x) has changed sign in between 0 to 1, so at some point x1 e (0,1) where g(x1) = 0

let x2 = x1 +1

then g(x1) =0 => f(x1+1) - f(x1) = 0 So f(x2) = f(x1)

(b) let g(x) = f(x) -x

g(0) = f(0) being a subset of [0, 2]

i.e 0 <= f(0) <=2

g(2) = f(2) - 2 But 0<= f(2) <= 2

so -2 <= g(2) <= 0

So either g(0) has changed sign in [0,2] or g(0) =0 or g(2) =0

In any of the above 3 cases, there exists c e [0,2] such that g(c) =0

i.e f(c) = c

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