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

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

SOLUTION:

$\frac{f\left(h+3\right)-f\left(3\right)}{h}=\frac{\left[3\left(h+3\right)^2-4\left(h+3\right)+1\right]-\left[3\left(3\right)^2-4\left(3\right)+1\right]}{h}$

$\frac{f\left(h+3\right)-f\left(3\right)}{h}=\frac{3\left(h^2+6h+9\right)-4h-12+1-16}{h}$

$\frac{f\left(h+3\right)-f\left(3\right)}{h}=\frac{3h^2+18h+27-4h-12+1-16}{h}$

$\frac{f\left(h+3\right)-f\left(3\right)}{h}=\frac{3h^2+14h}{h}$

$\frac{f\left(h+3\right)-f\left(3\right)}{h}=\frac{h\left(3h+14\right)}{h}$

$\frac{f\left(h+3\right)-f\left(3\right)}{h}=3h+14$

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