What is the equation of the line normal to # f(x)=sqrt(1/(x^2+2) # at # x=1#?

Answer 1

#y-sqrt(1/3)=root(3)9(x-1)#

in simplifying the equation we get

#(x^2+2)^(-1/2)#

We then apply the chain rule to get

#-x/(root(3)(x^2+2)^2#

Then we substitute in 1 for x to find the gradient of the tangent at x=1

This results us getting #-1/root(3)9#, however we need the normal's gradient, not the tangent's. The normal times the tangent gives us -1, so therefore the gradient of the normal will be #root(3)9#.
Now we know the gradient, we just need the co-ordinates which it runs through. We then sub x=1 back int the original equation to find the point which the normal will pass through. The y value, after subsitution is #sqrt(1/3)#.

Therefore by substituting the numbers/values into the point gradient formula, we'll find the equation of the normal. The point gradient formula is

#y-y_1=m(x-x_1)#
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Answer 2

The equation of the line normal to f(x) = sqrt(1/(x^2+2)) at x = 1 is y = -2x + 2.

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Answer from HIX Tutor

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.

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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