What is the local linearization of #F(x) = cos(x) # at a=pi/4?

Answer 1

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

#L(x)=f(a)+f'(a)(x-a)# #F(a)=cos (pi/4)=sqrt2/2# #F'(a)=-sin(pi/4)=-sqrt2/2# #L(x)=sqrt2/2-sqrt2/2 (x-pi/4)#
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Answer 2

To find the local linearization of ( F(x) = \cos(x) ) at ( a = \frac{\pi}{4} ), we first compute the value of ( F(a) ) and ( F'(a) ) at ( x = \frac{\pi}{4} ).

( F(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} )

( F'(\frac{\pi}{4}) = -\sin(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}} )

Now, using the point-slope form of the equation of a line ( y - y_1 = m(x - x_1) ), where ( (x_1, y_1) ) is a point on the line and ( m ) is the slope, we can write the local linearization as:

( L(x) = F(\frac{\pi}{4}) + F'(\frac{\pi}{4}) \cdot (x - \frac{\pi}{4}) )

Substituting the values of ( F(\frac{\pi}{4}) ) and ( F'(\frac{\pi}{4}) ):

( L(x) = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \cdot (x - \frac{\pi}{4}) )

( L(x) = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}x + \frac{1}{2} )

So, the local linearization of ( F(x) = \cos(x) ) at ( a = \frac{\pi}{4} ) is ( L(x) = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}x + \frac{1}{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.

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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