Question

How do you use the limit definition to find the slope of the tangent line to the graph `y = x^(3) + 2x^(2) - 3x + 2` at x=1?

Answer 1

See below.

Explanation:

We could use `lim_(hrarr0) (f(1+h)-f(1))/h` or `lim_(hrarr0) (f(x+h)-f(x))/h` then substitute `1` for `x`,

but I don't really want to get into cubing a binomial. So I'll use

`lim_(xrarr1)(f(x)-f(1))/(x-1) = lim_(xrarr1) ((x^3+2x^2-3x+2)-(2))/(x-1)`

`= lim_(xrarr1) (x^3+2x^2-3x)/(x-1)`

`= lim_(xrarr1)(x(x^2+2x-3))/(x-1)`

`= lim_(xrarr1)(x(x+3)(x-1))/(x-1)`

`= lim_(xrarr1)x(x+3)`

`= 4`