Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
f'\left( { - 1} \right) = \mathop {\lim }\limits_{x \to - 1} \frac{{f\left( x \right) - f\left( { - 1} \right)}}{{x - \left( { - 1} \right)}} = \mathop {\lim }\limits_{x \to - 1} \frac{{\left( {2{x^2} + 3x + 1} \right) - 0}}{{x + 1}} = \mathop {\lim }\limits_{x \to - 1} \frac{{2{x^2} + 3x + 1}}{{x + 1}} = \mathop {\lim }\limits_{x \to - 1} \frac{{\left( {x + 1} \right)\left( {2x + 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to - 1} \left( {2x + 1} \right) = 2.\left( { - 1} \right) + 1 = - 1\\
b,\\
f'\left( { - 1} \right) = \mathop {\lim }\limits_{x \to - 1} \frac{{f\left( x \right) - f\left( { - 1} \right)}}{{x - \left( { - 1} \right)}} = \mathop {\lim }\limits_{x \to - 1} \frac{{\left( {{x^3} - 3{x^2} + 2} \right) + 2}}{{x + 1}} = \mathop {\lim }\limits_{x \to - 1} \frac{{{x^3} - 3{x^2} + 4}}{{x + 1}} = \mathop {\lim }\limits_{x \to - 1} \frac{{\left( {x + 1} \right){{\left( {x - 2} \right)}^2}}}{{x + 1}} = \mathop {\lim }\limits_{x \to - 1} {\left( {x - 2} \right)^2} = {\left( { - 1 - 2} \right)^2} = 9\\
c,\\
f'\left( 1 \right) = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {3{x^2} - 4x + 9} \right) - 8}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{3{x^2} - 4x + 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {3x - 1} \right)\left( {x - 1} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {3x - 1} \right) = 3.1 - 1 = 2
\end{array}\)
