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How do you find the linearization of #F(x) = cos(x)# at a=pi/4?

Answer 1

Luke Alvarez

#L(x)=sqrt2/2-sqrt2/2(x-pi/4)#

The linearization at #x=a# is given by

#L(x)=f(a)+f'(a)(x-a)#

Knowing #f(x)=cosx, a=pi/4,# then

#f(pi/4)=cos(pi/4)=sqrt2/2#

#f'(x)=-sinx, f'(pi/4)=-sin(pi/4)=-sqrt2/2#

Our linearization is then

#L(x)=sqrt2/2-sqrt2/2(x-pi/4)#

Further simplification would not necessarily result in a cleaner expression.

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Answer 2

William Abdalla

To find the linearization of ( F(x) = \cos(x) ) at ( a = \frac{\pi}{4} ), use the formula for linearization:

[ L(x) = f(a) + f'(a)(x - a) ]

First, find ( f(a) ) and ( f'(a) ):

[ f\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} ]

[ f'(x) = -\sin(x) ] [ f'\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} ]

Now, plug these values into the linearization formula:

[ L(x) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) ] [ L(x) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}x + \frac{\pi}{4\sqrt{2}} ]

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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.