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How do you find the linearization at a=pi/4 of # f(x) = cos^(2)(x)#?

Answer 1

Samantha Adkison

the linearization of f(x) is #y=-x+pi/4+1/2#

the solution:

Given: #f(x)=cos^2 x# at #a=pi/4#

#f(a)=cos^2 a# #f(pi/4)=cos^2 (pi/4)=(1/sqrt2)^2=1/2#

We now have the point #(a, f(a))=(pi/4, 1/2)#

Compute for the slope #f'(a)#

#f' (x)=2*(cos x)(-sin x)#

#f' (a)=2*(cos a)(-sin a)#

#f' (pi/4)=2*(cos (pi/4))(-sin (pi/4))#

#f' (pi/4)=2*(1/sqrt2)(-1/sqrt2)=-1#

the linearization of #f(x)#

#y-f(a)=f' (a)(x-a)#

#y-1/2=-1(x-pi/4)#

#y=1/2-x+pi/4#

#y=-x+pi/4+1/2#

God bless....I hope the explanation is useful.

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Answer 2

Cameron Alexander

To find the linearization of ( f(x) = \cos^2(x) ) at ( a = \frac{\pi}{4} ), follow these steps:

Find the first derivative of ( f(x) ).
Evaluate the derivative at ( a = \frac{\pi}{4} ) to find the slope of the tangent line.
Use the point-slope form of a line with the slope and the point ( \left( \frac{\pi}{4}, f\left(\frac{\pi}{4}\right) \right) ) to find the equation of the tangent line, which represents the linearization.

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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.