Question

Limit definition for slope of the graph, equation of tangent line point for

f(x)=2x^2 at x=(-1)

Answer 1

The slope of the tangent line to
`f` at
`x=-1` is given by the derivative of
`f` at that point:

`f'(-1)=\displaystyle\lim_(x\to-1)(f(x)-f(-1))/(x-(-1))=\lim_(x\to-1)(2x^2-2)/(x+1)`

Factorize the numerator:

`2x^2-2=2(x^2-1)=2(x-1)(x+1)`

We have
`x` approaching -1; in particular, this means
`x\\eq-1`, so that

`(2x^2-2)/(x+1)=(2(x-1)(x+1))/(x+1)=2(x-1)`

Then

`f'(-1)=\displaystyle\lim_(x\to-1)(2x^2-2)/(x+1)=\lim_(x\to-1)2(x-1)=2(-1-1)=-4`

and the tangent line's equation is

`y-f(-1)=f'(-1)(x-(-1))\implies y-4x-2`