Let f(x)=−5.9sin(x)+1.6cos(x). What is the maximum and minimum value of this function?

1 Answer
Feb 23, 2018

See below.

Explanation:

The method I would use is to calculus.

If we find the first derivative of #-59/10sinx+8/5cosx#

#dy/dx(-59/10sinx+8/5cosx)=-59/10cosx-8/5sinx#

We know that Maximum/minimum and points of inflection occur where the first derivative is zero.

So solving for zero:

#-59/10cosx-8/5sinx=0#

multiply by #10#:

#-59cosx-16sinx=0#

#-59cosx=16sinx#

Divide by #-59cosx#:

#-(16sinx)/(59cosx)=1#

#sinx/cosx=-59/16#

#:.#

#tanx=-59/16#

#x=arctan(tanx)=arctan(-59/16)+pik#

#x=arctan(tanx)=pi +arctan(-59/16)+pik#

maximum values.

#-59/10sin(arctan(-59/16))+8/5cos(arctan(-59/16))~~6.1131#

minimum values.

#-59/10sin(pi+arctan(-59/16))+8/5cos(pi+arctan(-59/16))~~-6.1131#

So we get the maximum value #~~6.1131# for:

#arctan(-59/16)+2pik#

And we get the minimum value #~~-6.1131# for:

#pi+arctan(-59/16)+2pik#

#k in ZZ#

Note:

Normally when finding max/min values using derivatives, we would use the second derivative test to determine whether a point was a minimum or maximum. We know that this function is periodic, so once we found the zero gradients using the first derivative, we knew that this would just keep repeating at regular intervals. I don't know whether you have posted this question or not, but there may be an easier method than this, that doesn't involve calculus.

Graph:enter image source here