Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 9

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PROBLEM:

If f(x)=3x2−4x+1\displaystyle f\left(x\right)=3x^2-4x+1, find f(h+3)−f(3)h, h≠0\displaystyle \frac{f\left(h+3\right)-f\left(3\right)}{h},\:h\ne 0.

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SOLUTION:

f(h+3)−f(3)h=[3(h+3)2−4(h+3)+1]−[3(3)2−4(3)+1]h=3(h2+6h+9)−4h−12+1−16h=3h2+18h+27−4h−12+1−16h=3h2+14hh=h(3h+14)h=3h+14  (Answer)\begin{align*} \frac{f\left(h+3\right)-f\left(3\right)}{h} & =\frac{\left[3\left(h+3\right)^2-4\left(h+3\right)+1\right]-\left[3\left(3\right)^2-4\left(3\right)+1\right]}{h} \\ & =\frac{3\left(h^2+6h+9\right)-4h-12+1-16}{h}\\ & =\frac{3h^2+18h+27-4h-12+1-16}{h}\\ & =\frac{3h^2+14h}{h}\\ & =\frac{h\left(3h+14\right)}{h}\\ & =3h+14 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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