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) =

__ Ö13

×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 => 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

Thanks!