How many zeros does ( f(x)=cos(pi x) ) have in the interval ( I=[10,12] ) ?
Maple 绘图
A sensible guess might be . How do you know there are not more? We can use the IVT and monotonic functions to find out.
We would like to use the IVT on intervals where ( f(x) ) is monotonic. The function ( f ) is not monotonic on ( I ), but we can break ( I ) into two subintervals ( I1cup I2=I ) where ( f ) is monotonic on ( I1 ) and separately monotonic on ( I2 ).
Since ( f'(x)= ) , the function is decreasing on the interval ( I1= ) , increasing on the interval ( I2= ) .
We apply the IVT for ( f ) on these intervals separately. Since ( f(x) ) is continuous on the interval ( I1 ) and
( f(10)= ) ,
( f(11)= ) ,
then by the IVT there is at at least one zero in this interval. Since the function is monotonic on ( I1 ) there is exactly one zero in the interval ( I1 ).
Similarly, ( f(x) ) is continuous on the interval ( I2 ) and
( f(11)= ) ,
( f(12)= ) ,
.
Then by the IVT there is at least one zero in ( I2 ). Since the function is monotonic on ( I2 ) there is exactly one zero in ( I2 ). Hence the number of zeros in the interval ( I=I1cup I2 ) is exactly as we expected.
Note: the Maple syntax for the interval ( [1,pi ] ) is [1,Pi].