How do use the first derivative test to determine the local extrema #y = sin x cos x#?

2 Answers
Aug 1, 2015

Answer:

The extrema for #y=sin(x)cos(x)# are
#x=pi/4+npi/2#
with #n# a relative integer

Explanation:

Be #f(x)# the function representing the variation of #y# with repsect to #x#.

Be #f'(x)# the derivative of #f(x)#.

#f'(a)# is the slope of the #f(x)# curve at the #x=a# point.

When the slope is positive, the curve is increasing.
When the slope is negative, the curve is decreasing.
When the slope is null, the curve remains at the same value.

When the curve reaches an extremum, it will stop increasing/decreasing and start decreasing/increasing. In other words, the slope will go from positive to negative -or negative to positive- passing by the zero value.

Therefore, if you're looking for a function's extrema, you should look for its derivative's null values.

N.B. There is a situation when the derivative is null but the curve doesn't reach an extremum: it's called an inflection point. the curve will momentarily stop increasing/decreasing and then resume its increasing/decreasing. So you should also check if the slope's sign changes around its null value.

Example: #f(x)=sin(x)cos(x)=y#

#f'(x)=(dsin(x))/dxcdotcos(x)+sin(x)cdot(dcos(x))/dx#

#=cos(x)cdotcos(x)+sin(x)cdot(-sin(x))=cos^2(x)-sin^2(x)#

Now that we have the formula for #f'(x)#, we will look for its null values:

#f'(x)=cos^2(x)-sin^2(x)=0 rarr cos^2(x)=sin^2(x)#

The solutions are #pi/4+npi/2# with #n# a relative integer.

Aug 1, 2015

Answer:

Even if we plan to use the first derivative test, it is worth observing that #y = 1/2 sin(2x)#.

Explanation:

Having made that observation, we do not really need calculus to find the extrema.

We can rely on our knowledge of trigonometry and the graphs of sinusoidal functions

The maximum value (of 1/2) will occur when #2x = pi/2 + 2pik# or when #x = pi/4 + pik# for #k# an integer.

The minimum occurs at #x = 3pi/4 + pik# for #k# an integer.

We can use the derivative, but we do not really need it.

Using the Derivative

Having rewritten #y#, we can quickly see that #y' = cos(2x)#

So the critical numbers for #y# are #2x = pi/2 + 2pik# and #2x = (3pi)/2 + 2pik#, (when the cosine is #0#) or

#x= pi/4+pik# and #x= (3pi)/4+pik#

Checking the sign of #y ' = cos(2x)#, we will find maximum values at the first set of critical numbers and minimum values at the second.