How do I find the linear approximation of the function #g(x)=(1+x)^(1/3)# at #a=0#?

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Answer 1 · 2018-04-09T16:11:12

We have

#g(a) = g(0) = (1 +0)^(1/3) = 1#

Now taking the derivative.

#g'(x) = 1/3(1 + x)^(-2/3)#
#g'(a) = g'(0) = 1/3(1 + 0)^(-2/3) = 1/3#

Now we find the equation of the tangent.

#y -y_1 = m(x - x_1)#

#y - 1 = 1/3(x - 0)#

#y = 1/3x + 1#

As you can see this approximates the function relatively well for value of #a# in the region of #0#.

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Hopefully this helps!