Question

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)`

Explanation:

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)`