# How do you find the equation of a line tangent to the function #y=1+x^(2/3)# at (0,1)?

The general slope of a line is given by it's derivative.

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To find the equation of a line tangent to the function y=1+x^(2/3) at (0,1), we need to find the slope of the tangent line at that point.

To find the slope, we can take the derivative of the function y=1+x^(2/3) with respect to x.

The derivative of y=1+x^(2/3) is dy/dx = (2/3)x^(-1/3).

Substituting x=0 into the derivative, we get dy/dx = (2/3)(0)^(-1/3) = undefined.

Since the derivative is undefined at x=0, it means that the tangent line is vertical.

The equation of a vertical line passing through the point (0,1) is x=0.

Therefore, the equation of the line tangent to the function y=1+x^(2/3) at (0,1) is 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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