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

Question

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

Elijah Abram · Accepted Answer

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

Emily Ammerman · Answer

undefined To find the local maximum value of $ f $ using the first and second derivative tests for $ f(x) = \sin(x) $:

1. Find the first derivative: $ f'(x) = \cos(x) $.
2. Set $ f'(x) $ equal to 0 to find critical points: $ \cos(x) = 0 $.
3. Solve for $ x $: $ x = \frac{\pi}{2} + \pi n $, where $ n $ is an integer.
4. Find the second derivative: $ f''(x) = -\sin(x) $.
5. Test the critical points in the second derivative test:
   - At $ x = \frac{\pi}{2} $, $ f''(\frac{\pi}{2}) = -\sin(\frac{\pi}{2}) = -1 $, which is less than 0.
   - Thus, $ f $ has a local maximum at $ x = \frac{\pi}{2} $.
6. Calculate the local maximum value: $ f(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) = 1 $.

Therefore, the local maximum value of $ f(x) = \sin(x) $ is 1 at $ x = \frac{\pi}{2} $.