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How do you find the linearization of #f(x)=cosx# at x=5pi/2?

Answer 1

Ezra Adams

#cosx~=(5pi)/2-x#

The linear approximation of a differentiable function around #x=barx# is given by the tangent line, so that:

#f(x)~=f(barx)+f'(barx)(x-barx)#

For #f(x) = cosx# at #barx= (5pi)/2#.

#cosx~=cos((5pi)/2) -sin((5pi)/2)(x-(5pi)/2)#

Considering that:

#(5pi)/2 = 2pi+pi/2#, #cos(x+2pi)= cosx# #sin(2pi+x) = sinx# #cos(pi/2)=0# #sin(pi/2)=1#

This becomes:

#cosx~=(5pi)/2-x#

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Answer 2

Elijah Abram

To find the linearization of ( f(x) = \cos(x) ) at ( x = \frac{5\pi}{2} ), follow these steps:

Find the value of ( f(x) ) at ( x = \frac{5\pi}{2} ). This will be the ( y )-coordinate of the point on the curve.
Find the derivative of ( f(x) = \cos(x) ).
Evaluate the derivative at ( x = \frac{5\pi}{2} ). This will be the slope of the tangent line.
Use the point-slope form of a line to write the equation of the tangent line using the point ( \left(\frac{5\pi}{2}, f\left(\frac{5\pi}{2}\right)\right) ) and the slope obtained in step 3.

The linearization ( L(x) ) of ( f(x) ) at ( x = \frac{5\pi}{2} ) will be the equation of the tangent line.

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Answer from HIX Tutor

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