Show that for any real constants c and d, the eqation x = c + dcos(x) has at least one root.

My professor gave the hint: Find an interval [a,b] on which

f(x) = x - c - dcos(x) changes sign.

Any advice?

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- September 27th 2009, 09:14 PMjzelltRoot
Show that for any real constants c and d, the eqation x = c + dcos(x) has at least one root.

My professor gave the hint: Find an interval [a,b] on which

f(x) = x - c - dcos(x) changes sign.

Any advice? - September 27th 2009, 10:17 PMCalculus26
f(x) = x - c - dcos(x)

__Case 1 dcos(c) > 0__

f(c) = -dcos(c) < 0

f(c + pi) = pi -dcos(c+pi) = pi + dcos(c) > 0

We're done

__Case 2 dcos(c) < 0__

f(c) = -dcos(c) > 0

Let n be such that npi > dcos(c)

f(c-npi) = -npi -dcos(c-npi) = -npi - dcos(c)cos(npi)

cos(npi) =__+__1

in either case f(c-npi) < 0