Subtracting a^3 and canceling h from the numerator and denominator gives us the final answer. (Simplify your answer completely.) (f(a+h)-f(a))/h =
To simplify the expression (f(a+h)-f(a))/h, we need to first expand f(a+h) using the binomial expansion. However, since we do not have the specific function f(a), we cannot proceed further with simplification.
To simplify the expression (f(a+h)-f(a))/h, we can expand f(a+h) using the binomial theorem.
Let's assume f(x) is a function.
According to the binomial theorem, we can express f(a+h) as:
f(a+h) = f(a) + f'(a)h + (f''(a)/2!)h^2 + (f'''(a)/3!)h^3 + ...
Now we substitute this expansion into the expression:
[(f(a) + f'(a)h + (f''(a)/2!)h^2 + (f'''(a)/3!)h^3 + ...) - f(a)] / h
Expanding further:
[f(a) - f(a)] + [f'(a)h - f(a)] + [(f''(a)/2!)h^2 - f(a)] + [(f'''(a)/3!)h^3 - f(a)] + ...
The first term "f(a) - f(a)" cancels out to zero. So we are left with:
[f'(a)h - f(a)] + [(f''(a)/2!)h^2 - f(a)] + [(f'''(a)/3!)h^3 - f(a)] + ...
Now we can factor out the common factor of -f(a):
-f(a) + f'(a)h + (f''(a)/2!)h^2 + (f'''(a)/3!)h^3 + ...
Finally, we divide the entire expression by h:
(-f(a) + f'(a)h + (f''(a)/2!)h^2 + (f'''(a)/3!)h^3 + ...) / h
Now we can cancel out h from the numerator:
-f(a)/h + f'(a) + (f''(a)/2!)h + (f'''(a)/3!)h^2 + ...
Thus, the final answer is:
-f(a)/h + f'(a) + (f''(a)/2!)h + (f'''(a)/3!)h^2 + ...