Hi,
I did a trig problem today and was not sure how to get the smallest
positive value of x when it occurs. You'll see what I mean
below.
After doing part (a) of the question I got
2cos4x + 3sin4x = Rcos(4x - t) = sqrt(13)×Cos(4x - 0.98)
(R>0, t lies in the interval [0,p/2])
The question asked me to work out the Minimum of
(1/(2cos4x+3sin4x)) and the smallest positive value of x when this
occurs.
Help appreciated.
To consider maxima and minima of your function first consider
the function cos(x). As -1£cos(x)£1
clearly the function f(x)=sqrt(13)cos(4x-0.98) will lie between
-sqrt(13) and +sqrt(13) (this is because cos(4x-0.98) will still
lie between -1 and +1). So to find the minimum value of 1/f(x) we
must find the maximum value of f(x). This will be when cos(4x-0.98)
is equal to 1 so max of f(x)=sqrt(13), so the minimum of 1/f(x)
will be 1/sqrt(13). To find the smallest positive value of x when
this is true you must solve the equation;
cos(4x-0.98)=1
so 4x-0.98=0+2np where n is an integer,
this is because cos has period 2p , so
as the smallest postive value is required this will happen when
n=0.
So 4x=0.98 so x=0.245698.. (be careful you don't use your rounded
down figure of tan-1(3/2) in the calculation).
Thanks Narendra!
Can't the min be -1/sqrt(13)? or am I wrong?
The x=0.245.. is great!
Be careful! There are no absolute minimum
for the expression 1/(2 cos 4x + 3 sin 4x), as we can make the
denominator as close to zero as we like, and yet less than
zero.
I'd agree that the question is awkwardly phrased. It would be
better if they asked for the local minimum of the expression for
1/(2 cos 4x + 3 sin 4x), but these terminology unfortunately will
put people off....
Kerwin
Kerwin,
What does 'local minimum' mean. Is that just meaning the nearest
min to the origin?
Local minimum is where a function attains
a minimum with respect to its neighbourhood (i.e. dy/dx=0, and a
small change in x causes an increase in y), whereas absolute
minimum is where the smallest value is attained.
For example, consider the function
f(x)=x3-x, -2£x£2
The absolute max. and min actually occurs at x=2 and x=-2
respectively.(Draw a graph if you are not convinced).
However...
f'(x)=3x2-1=0 Þ
x=±1/sqrt(3)
and it is clear from the graph of f that x=1/sqrt(3) corresponds to
a local minimum, and x=-1/sqrt(3) corresponds to a local
maximum.
Kerwin