Answer:
See the answers below.
Step-by-step explanation:
[tex]a.\:\frac{f\left(x\right)-f\left(a\right)}{x-a}=\frac{2x^2-x-5-\left(2a^2-a-5\right)}{x-a}\\\\=\frac{2x^2-x+a-2a^2}{x-a}\\\\=\frac{2\left(x+a\right)\left(x-a\right)-1\left(x-a\right)}{x-a}\\\\=\frac{\left(x-a\right)\left[2\left(x+a\right)-1\right]}{x-a}\\\\=2x+2a-1\\\\\\b.\:\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{2\left(x+h\right)^2-\left(x+h\right)-5-\left(2x^2-x-5\right)}{h}\\\\=\frac{2\left(x^2+2xh+h^2\right)-\left(x+h\right)-5-\left(2x^2-x-5\right)}{h}\\[/tex]
Expand and simplify to get:
[tex]=\frac{2h^2+4xh-h}{h}\\\\=\frac{h\left(2h+4x-1\right)}{h}\\\\=2h+4x-1[/tex]
Best Regards!
