Answer
$$\frac{f(x+h) - f(x)}{h} = 4x + 2h - 1$$
Work Step by Step
To answer this exercise, one must first find $f(x + h)$:
$$f(x) = 2x^{2} - x - 1$$
Therefore,
$$f(x + h) = 2(x + h)^{2} - (x+h) - 1$$
$$f(x + h) = 2(x^{2} + 2xh + h^{2}) - (x + h) - 1$$
$$f(x + h) = 2x^{2} + 4xh + 2h^{2} - x - h - 1$$
Finally, one simply substitutes the needed values to solve for $\frac{f(x+h) - f(x)}{h}$:
$$\frac{f(x+h) - f(x)}{h} = \frac{(2x^{2} + 4xh + 2h^{2} - x - h - 1) - (2x^{2} - x - 1)}{h}$$
$$\frac{f(x+h) - f(x)}{h} = \frac{2x^{2} + 4xh + 2h^{2} - x - h - 1 - 2x^{2} + x + 1}{h}$$
$$\frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^{2} - h }{h}$$
$$\frac{f(x+h) - f(x)}{h} = \frac{h(4x + 2h - 1)}{h}$$
$$\frac{f(x+h) - f(x)}{h} = 4x + 2h - 1$$
