Question

How do you use the limit definition to find the slope of the tangent line to the graph `f(t) = t − 13 t^2` at t=3?

Answer 1

`f'(3)=-77`

Explanation:

The slope of the tangent at any point is given by the derivative of the function at that point. The definition of the derivative of `y=f(x)` is

`f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`

So with `f(t) = t-13t^2` and the value of `f'(3)` sought then;

`f'(3)=lim_(h rarr 0) (f(3+h)-f(3))/h`
`\ \ \ \ \ \ \ \ \=lim_(h rarr 0) ({(3+h)-13(3+h)^2}-{3-13*3^2})/h`

`\ \ \ \ \ \ \ \ \=lim_(h rarr 0) ({3+h-13(9+6h+h^2)}-{3-13*9})/h`

`\ \ \ \ \ \ \ \ \=lim_(h rarr 0) ({3+h-117-78h-13h^2}-{3-117})/h`

`\ \ \ \ \ \ \ \ \=lim_(h rarr 0) {3+h-117-78h-13h^2-3+117)/h`

`\ \ \ \ \ \ \ \ \=lim_(h rarr 0) {h-78h-13h^2)/h`

`\ \ \ \ \ \ \ \ \=lim_(h rarr 0) (-77-13h)`

`\ \ \ \ \ \ \ \ \=-77`