How do you find the critical numbers of #f(x)=sinxcosx#?
Critical numbers of
# x \ \ = pi/4 + (npi)/2 \ \ \ \ n in ZZ#
eg#x = +pi/4, +3pi/4, +5pi/4, ... #
We have:
# f(x) = sinx cos x #
Differentiating wrt
# f'(x) = (sinx) (sinx) + (cosx) (cosx) #
# " "= cos^2x sin^2x #
# " "= cos(2x) #
At a critical point
# :. cos (2x) = 0 #
# :. 2x = pi/2 + npi #
# :. x \ \ = pi/4 + (npi)/2 \ \ \ \ n in ZZ#
Hence critical numbers of
# x \ \ = pi/4 + (npi)/2 \ \ \ \ n in ZZ#
eg#x = +pi/4, +3pi/4, +5pi/4, ... #
We can see these values of
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To find the critical numbers of ( f(x) = \sin(x)\cos(x) ), you need to find the values of ( x ) where the derivative ( f'(x) ) is equal to zero or undefined.

Calculate the derivative ( f'(x) ): [ f'(x) = \cos(x)\cos(x)  \sin(x)\sin(x) ] [ f'(x) = \cos^2(x)  \sin^2(x) ]

Use the trigonometric identity ( \cos^2(x)  \sin^2(x) = \cos(2x) ): [ f'(x) = \cos(2x) ]

Set ( f'(x) ) to zero and solve for ( x ): [ \cos(2x) = 0 ]

Solve for ( x ): [ 2x = \frac{\pi}{2} + n\pi ] where ( n ) is an integer.
[ x = \frac{\pi}{4} + \frac{n\pi}{2} ]
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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