Answer
An approximation for $f'(1) = 0.003001$
The actual value of $f'(1) = 0$
The two values are comparable. Therefore, we confirm both the approximation and the actual value of $f'(1)$
Work Step by Step
We approximate $f'(1)$ using the difference quotient $\frac{f(1+h)-f(1)}{h}$ for values of $h$ near $0$. In this case, we choose $h=0.001$.
Approximate value of $f'(1) = \frac{f(1+h)-f(1)}{h}=\frac{f(1+0.001)-f(1)}{0.001} = \frac{(1.001^3-3(1.001)+1)-(1^3 - 3(1) + 1)}{0.001} = 0.003001$
We apply the power rule to take the derivative and find the real value $f'(1)$:
Real value of $f'(1) = 3x^2-3 = 3(1^2)-3 = 0$
