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

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Answer 1 · 2016-08-26T17:42:25

See below.

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

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

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

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

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

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

$= 4$

How do you use the limit definition to find the slope of the tangent line to the graph y = x^(3) + 2x^(2) - 3x + 2y=x3+2x2−3x+2y = x^(3) + 2x^(2) - 3x + 2 at x=1?