How do you find the linearization of #f(x) = x^(1/2) # at x=25?

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Answer 1 · 2016-11-23T19:33:29

The linearization is the tangent line. (Or maybe it is more helpful to say: it is a way of thinking about and using the tangent line.)

#f(x) = x^(1/2) = sqrtx#

At #x=25#, we have #y = f(25) = 5#

#f'(x) = 1/(2sqrtx)# so at #x=25#, the slope of the tangent line is #m=f'(25) = 1/10#

Equation of tangent line in point-slope form:

#y-5 = 1/10(x-25)#

Linearization at #a=1# (in a form I am used to):

#L(x) = 5+1/10(x-25)#

Free example of using the linearization

#sqrt28 = f(28)#

#f(28) ~~ L(28) = 5+1/10(28-25) = 5+1/10(3) = 5+0.3 = 5.3#

(The point on the tangent line with #x#-coordinate #28# has #y#-coordinate #5.3#.)