Answer:
[tex]f'(x) = 3-2\cdot x[/tex]
Step-by-step explanation:
Let be [tex]f(x) = -x^{2}+3\cdot x[/tex], such that:
[tex]f'(x) = \frac{f(x+h)-f(x)}{h}[/tex]
Hence,
[tex]f(x+h) = -(x+h)^{2}+3\cdot (x+h)[/tex]
[tex]f(x+h) = -(x^{2}+2\cdot x\cdot h + h^{2})+3\cdot x + 3\cdot h[/tex]
[tex]f(x+h) = -x^{2}+(3-2\cdot x) \cdot h - h^{2}+3\cdot x[/tex]
[tex]f(x) = -x^{2}+3\cdot x[/tex]
Then,
[tex]f'(x) = \frac{-x^{2}+(3-2\cdot x)\cdot h-h^{2}+3\cdot x+x^{2}-3\cdot x}{h}[/tex]
[tex]f'(x) = \frac{(3-2\cdot x)\cdot h-h^{2}}{h}[/tex]
[tex]f'(x) = 3-2\cdot x -h[/tex]
If [tex]h \longrightarrow 0[/tex], then:
[tex]f'(x) = 3-2\cdot x[/tex]