How do I find the linear approximation of the function #g(x)=(1+x)^(1/3)# at #a=0#?
1 Answer
Apr 9, 2018
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
Hopefully this helps!
