How do you use the intermediate value theorem to explain why #f(x)=x^2-x-cosx# has a zero in the interval [0,pi]?

1 Answer
Jan 8, 2017

Answer:

See explanation...

Explanation:

The intermediate value theorem tells us that if #f(x)# is a function continuous on an interval #[a, b]# and #y in [f(a), f(b)]# (or #y in [f(b), f(a)]# if #f(b) < f(a)#), then there is some #x in [a, b]# such that #f(x) = y#.

In our example:

#f(x) = x^2-x-cos x#

#f(x)# is continuous on #[0, pi]# (in fact on the whole of #RR#)

#f(0) = 0-0-cos(0) = -1 < 0#

#f(pi) = pi^2-pi-cos(pi) = pi(pi-1)+1 > 3(3-1)+1 = 7 > 0#

So #0 in [f(0), f(pi)]# and by the intermediate value theorem, there is some #x in [0, pi]# such that #f(x) = 0#