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

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Answer 1 · 2017-02-05T17:10:02

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#