# What is the slope of the line normal to the tangent line of #f(x) = x-sqrt(x^2+4) # at # x= 2 #?

Slope of normal is

graph{((sqrt2-1)x-sqrt2y-2)(sqrt2x-y(1-sqrt2)-6sqrt2+6)(x-sqrt(x^2+4)-y)=0 [-5, 5, -2.5, 2.5]}

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To find the slope of the line normal to the tangent line of ( f(x) = x - \sqrt{x^2 + 4} ) at ( x = 2 ), we first need to find the derivative of the function ( f(x) ) and evaluate it at ( x = 2 ) to get the slope of the tangent line. Then, we can find the negative reciprocal of this slope to obtain the slope of the line normal to the tangent line.

The derivative of ( f(x) ) can be found using the chain rule: [ f'(x) = 1 - \frac{x}{\sqrt{x^2 + 4}} ]

Evaluate ( f'(x) ) at ( x = 2 ): [ f'(2) = 1 - \frac{2}{\sqrt{2^2 + 4}} = 1 - \frac{2}{\sqrt{8}} = 1 - \frac{2}{2\sqrt{2}} = 1 - \frac{1}{\sqrt{2}} ]

This gives us the slope of the tangent line at ( x = 2 ).

The negative reciprocal of this slope will give us the slope of the line normal to the tangent line: [ \text{Slope of normal line} = -\frac{1}{f'(2)} = -\frac{1}{1 - \frac{1}{\sqrt{2}}} ]

Simplify the expression: [ = -\frac{1}{1 - \frac{1}{\sqrt{2}}} \times \frac{\sqrt{2}}{\sqrt{2}} ] [ = -\frac{\sqrt{2}}{\sqrt{2} - 1} \times \frac{\sqrt{2}}{\sqrt{2}} ] [ = -\frac{2}{2 - \sqrt{2}} ]

So, the slope of the line normal to the tangent line of ( f(x) ) at ( x = 2 ) is ( -\frac{2}{2 - \sqrt{2}} ).

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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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