How do you find the critical points for # f(x)=3cos2(x-pi/4)+1#?
Use trigonometry to simplify the expression for
Without rewriting we can get the same answer (of course).
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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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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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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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