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

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!

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To find the linear approximation of the function ( g(x) = (1+x)^{\frac{1}{3}} ) at ( a = 0 ), follow these steps:

- Find the first derivative of ( g(x) ), denoted ( g'(x) ).
- Evaluate ( g'(0) ) to find the slope of the tangent line at ( x = 0 ).
- Use the point-slope form of a line to write the equation of the tangent line.

Let's go through each step:

- The first derivative of ( g(x) ) is ( g'(x) = \frac{1}{3}(1+x)^{-\frac{2}{3}} ).
- Evaluate ( g'(0) ): ( g'(0) = \frac{1}{3}(1+0)^{-\frac{2}{3}} = \frac{1}{3} ).
- Use the point-slope form of a line with the point ( (0, g(0)) = (0, 1) ) and slope ( g'(0) = \frac{1}{3} ):

[ y - y_1 = m(x - x_1) ] [ y - 1 = \frac{1}{3}(x - 0) ] [ y - 1 = \frac{1}{3}x ]

This is the equation of the tangent line at ( x = 0 ).

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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.

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.

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.

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