Tangent/first principles

[bully3000]

Consider the function f(x)=x^2-1. Find the equation of the tangent to the graph of f(x) at x=3. When calculating f'(3), use first principles.

Any help would be appreciated!

[Lasercookie]

Consider the function f(x)=x^2-1. Find the equation of the tangent to the graph of f(x) at x=3. When calculating f'(3), use first principles.

Any help would be appreciated!

So our definition of first principles is . I'm assuming you're familiar with the geometric interpretation of this, and the intuition for why this is the tangent. If not read this https://en.wikipedia.org/wiki/Derivative#Definition_via_difference_quotients, or textbook, or khanacademy etc.

So let's find . Once we sub our stuff into that quotient it's really just a bit of algebra to simplify it. The annoying bit is what appears to be a h -> 0 in the dominator.



I'd expect you to be able to do the rest of the algebra yourself. Here's my working out, but don't look till you do it yourself.

Spoiler

since when we have

So that means that We can see quite easily to be true if we differentiate the usual way we'd do it.

To find the equation of the tangent at x = 3, calculate the gradient at x =3 (f'(3)) and the values of the coordinate (3, f(3)). Then sub it into our relationship for a linear line (y = mx+c and what not).

So just evaluate and simplify