How do you find local maximum value of f using the first and second derivative tests: #f(x)= sinx#?

1 Answer
Nov 15, 2016

Answer:

The max occurs at #x=pi/2# where #f(x)=1#

Explanation:

#1.# The 1st derivative test: #f'(x)=cos x#

#cos x=0# at #x= pi/2, (3pi)/2, "etc."#

Make a sign chart for #f'(x)# with a critical point at #pi/2#. To the left of #pi/2#, #cos x# is positive; to the right, it is negative. Therefore, there is a max at #pi/2#

#2.# The 2nd derivative test: #f^('')(x) = - sin (x)#

At the critical point where #x=pi/2#, you have

#-sin(pi/2) = -1 <0#

Since the second derivative is negative at #x=pi/2#, the curve is concave down at that point. Therefore, it is a max.