Answer:
8+h
Step-by-step explanation:
Simply substitute x = 2+h and x = 2 in.
[tex]\displaystyle \large{\frac{f(2+h)-f(2)}{h}=\frac{(2+h)^{2} +4(2+h)+5-[(2)^2+4(2)+5]}{h}}\\\displaystyle \large{\frac{f(2+h)-f(2)}{h}=\frac{4+4h+h^2 +8+4h+5-17}{h}}\\\displaystyle \large{\frac{f(2+h)-f(2)}{h}=\frac{h^2 +8h}{h}=h+8}[/tex]
Therefore, the answer is h+8 or 8+h.
Another way is to differentiate the function with respect to x. This equation or expression is rate of changes from 2 to 2+h with h being any numbers.
The definition of derivative is:
[tex]\displaystyle \large{f\prime(x)= \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}}[/tex]
Notice how both f(2+h)-f(2) over h and the definition of derivative look same except the derivative has limit of h approaching to 0, making h not having any values by default.
Since f(2+h)-f(2) over h does not have limit, we have to add +h when differentiating the function.
[tex]\displaystyle \large{f\prime(x)=2x+4}\\\displaystyle \large{f\prime(2)=2(2)+4}\\\displaystyle \large{f\prime(2)=8}[/tex]
Since f’(2) is 8 but because f(2+h)-f(2)/h does not have limit of h, therefore we add +h.
[tex]\displaystyle \large{\frac{f(2+h)-f(2)}{h}=8+h}}[/tex]
