Answer
$3$
Work Step by Step
Consider the tangent line that contains the point $(x, c)$. The slope the tangent line to the graph of $f(x)$ at $(x, c)$ can be written as
$f'(x) =\lim\limits_{x \to c} \dfrac{f(x)-f(c)}{x-c}$
Consider $c=1$ and $f(x)=x^3+x^2-2x$. Thus, we have:
$f'(1) =\lim\limits_{x \to 1} \dfrac{f(x)-f(1)}{x-1} \\=\lim\limits_{x \to 1} \dfrac{(x^3+x^2-2x)-0}{x-1}\\=\lim\limits_{x \to 1} \dfrac{(x^2+2x)(x-1)}{x-1} \\=\lim\limits_{x \to 1} x^2+2x \\=3 $