Question #5022

Suppose f is continuous on all real numbers and f(c) is not equal to zero for some point c in all real numbers. Show that there exists an interval (a, b) containing c such that f is never 0 on this interval

Expert's answer

f(c) is not equal to zero. Let f(c)>0. Then there exists such epsilon>0:

f(c)>epsilon.

As f is continuous function in c, then for each

epsilon1>0 there exists delta>0 such that for each x:

if

abs(c-x)<delta, then abs(f(c)-f(x))<epsilon1.

Let epsilon1 = epsilon /

2. Then there exists delta1 such that:

abs(f(c)-f(x))<epsilon/2 for each x

from (c-delta1, c+delta1). As f(c)>epsilon, then f(x) > epsilon / 2 >

0.

Let a = c-delta1, b = c+delta1, (a,b) contains c. So, we get that for each

x from (a,b), f(x)>0.

The proof is the same for f(c)<0.

f(c)>epsilon.

As f is continuous function in c, then for each

epsilon1>0 there exists delta>0 such that for each x:

if

abs(c-x)<delta, then abs(f(c)-f(x))<epsilon1.

Let epsilon1 = epsilon /

2. Then there exists delta1 such that:

abs(f(c)-f(x))<epsilon/2 for each x

from (c-delta1, c+delta1). As f(c)>epsilon, then f(x) > epsilon / 2 >

0.

Let a = c-delta1, b = c+delta1, (a,b) contains c. So, we get that for each

x from (a,b), f(x)>0.

The proof is the same for f(c)<0.

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