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How do you find the linearization of #f(x) = x^(1/2)# at a=16?

Answer 1

Elijah Abram

The linearization is:

#y=x/8+2#

We can approximate a curve using its tangent line:

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

For #a=16# we have:

#f(a) = 16^(1/2)=sqrt(16) = 4#

#f'(x) = d/dx x^(1/2) = 1/(2sqrtx)#

#f'(a) = 1/(2sqrt16) = 1/8#

so the linarization is:

#y=4+1/8(x-16)#

#y=x/8+2#

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Answer 2

Adam Adkins

To find the linearization of ( f(x) = x^{1/2} ) at ( a = 16 ), follow these steps:

Find the derivative of ( f(x) ), denoted as ( f'(x) ).
Evaluate ( f'(a) ) at ( a = 16 ).
Write the equation of the tangent line using the point-slope form ( y - y_1 = m(x - x_1) ), where ( (x_1, y_1) ) is the point ( (16, f(16)) ) and ( m = f'(16) ).

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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.