Solución

Primero calculamos f′(x)f ′ (x)

f′(x)=lim⁡h→0(2(x+h)2−3(x+h)+1)−(2x2−3x+1)hSustiyendo   f(x)=2x3−3x+1   y   y   f(x+h)=lim⁡h→04xh+2h2−3hhSimplificando el denominador=lim⁡h→0(4x+2h−3)Sacando factor comuˊn   h   y simplicando=4x−3Calculando el lıˊmite \begin{aligned} f ′ (x) &= \mathop {\lim }\limits_{h \to 0} \frac{(2 (x + h)^2−3 (x + h) +1) - (2x^2−3x + 1)}{ h} &\text{Sustiyendo} \,\,\,f(x)=2x^3-3x+1\,\,\,\text{y}\,\,\,\text{y}\,\,\,f(x+h)\\ &= \mathop {\lim }\limits_{h \to 0} \frac{4xh + 2h^2−3h}{h } &\text{Simplificando el denominador}\\ &= \mathop {\lim }\limits_{h \to 0} (4x + 2h − 3) &\text{Sacando factor común}\,\,\,h\,\,\,\text{y simplicando}\\ &= 4x − 3 &\text{Calculando el límite} \end{aligned}

Ahora buscamos f′′(x)f '' (x) calculando la derivada de f′(x)=4x−3f ′ (x) = 4x − 3.

f′′(x))lim⁡h→0f′(x+h)−f′(x)hUtilizando   f′(x)=lim⁡h→0f′(x+h)−f′(x)h=lim⁡h→04(x+h)−3)−(4x−3)hSustituyendo   f′(x+h)=4(x+h)−3   y   f′(x)=4x−3=lim⁡h→04Simplificando=4Tomando lıˊmites\begin{aligned} f '' (x) &)\mathop {\lim }\limits_{h \to 0} \frac{f ′ (x + h) −f ′ (x)} {h} &\text{Utilizando}\,\,\,f'(x)=\mathop {\lim }\limits_{h \to 0} \frac{f'(x+h)-f'(x)}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{4 (x + h) −3) - (4x − 3)} {h} &\text{Sustituyendo}\,\,\,f'(x+h)=4(x+h)-3\,\,\, \\ & &\text{y}\,\,\,f'(x)=4x-3\\ &= \mathop {\lim }\limits_{h \to 0} 4 &\text{Simplificando}\\ &= 4 &\text{Tomando límites} \end{aligned}