How do you find the critical points for # f(x)=3cos2(x-pi/4)+1#?

Answer 1

Use trigonometry to simplify the expression for #f(x)#.

#cos2(x-pi/4) = cos(2x-pi/2) = sin2x#
(If you don't remember the identity above, use the difference formula for #cos(2x-pi/2) = cos(2x)cos(pi/2)+sin(2x)sin(pi/2) = sin(2x)#)
So #f(x) = 3sin(2x)+1#
and #f'(x) = 6cos(2x)# which exists for all #x# and is #0# when
#cos(2x) = 0#. Which is true exactly when
#2x = pi/2 + pik# for integer #k#. Or,
#x = pi/4 + pi/2k# for integer #k#.

Without rewriting we can get the same answer (of course).

#f(x) = 3cos2(x-pi/4)+1#
#f'(x) = -3sin2(x-pi/4)*[d/dx(2(x-pi/4)] = -3sin2(x-pi/4)*[2(1)]#
#f'(x) = -6sin2(x-pi/4)#
F'(x) is never undefined and it is #0# for
#sin2(x-pi/4) = 0# Which is true exactly when
#2(x-pi/4) = pik# for integer #k#. Or,
#x -pi/4= pi/2k# for integer #k#. So we need
#x = pi/4+ pi/2k# for integer #k#.
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Answer 2

To find the critical points of ( f(x) = 3\cos^2(x - \frac{\pi}{4}) + 1 ), you first need to find its derivative, then solve for ( x ) when the derivative equals zero.

Taking the derivative of ( f(x) ), we get:

( f'(x) = -6\cos(x - \frac{\pi}{4})\sin(x - \frac{\pi}{4}) )

Setting ( f'(x) ) equal to zero and solving for ( x ) gives us the critical points.

( -6\cos(x - \frac{\pi}{4})\sin(x - \frac{\pi}{4}) = 0 )

( \Rightarrow \cos(x - \frac{\pi}{4}) = 0 ) or ( \sin(x - \frac{\pi}{4}) = 0 )

Solving these equations will give us the critical points. The solutions will be ( x = \frac{\pi}{4} + k\pi ) and ( x = \frac{\pi}{4} + \frac{\pi}{2} + k\pi ), where ( k ) is an integer.

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Answer 3

To find the critical points of ( f(x) = 3\cos^2(x - \frac{\pi}{4}) + 1 ), we first need to find its derivative, ( f'(x) ), and then solve for ( x ) when ( f'(x) = 0 ) or ( f'(x) ) is undefined.

[ f'(x) = -6\cos(x - \frac{\pi}{4})\sin(x - \frac{\pi}{4}) ]

Setting ( f'(x) ) to zero and solving for ( x ):

[ -6\cos(x - \frac{\pi}{4})\sin(x - \frac{\pi}{4}) = 0 ]

This equation is true when either ( \cos(x - \frac{\pi}{4}) = 0 ) or ( \sin(x - \frac{\pi}{4}) = 0 ).

For ( \cos(x - \frac{\pi}{4}) = 0 ), we have:

[ x - \frac{\pi}{4} = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots ]

[ x = \frac{3\pi}{4}, \frac{7\pi}{4}, \frac{11\pi}{4}, \ldots ]

For ( \sin(x - \frac{\pi}{4}) = 0 ), we have:

[ x - \frac{\pi}{4} = 0, \pi, 2\pi, \ldots ]

[ x = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \ldots ]

Therefore, the critical points of ( f(x) ) are ( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \ldots ).

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Answer from HIX Tutor

When evaluating a one-sided 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 one-sided 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 one-sided 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 one-sided 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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