Answer
See explanation
Work Step by Step
We will use the following formulas frequently:
\begin{align}
\textbf{Formula (1):}\quad L(x)&=f(x_0)+f'(x_0)(x-x_0),\\
\textbf{Formula (2):}\quad L(x)&=f(x_0)+f'(x_0)\Delta x,\quad\Delta x:=x-x_0.
\end{align}
1. Approximations for $f(x)=x^3$ at $x_0=1$
{(a)} Let $f(x)=x^3$. Then
\[f'(x)=3x^2,\qquad f'(1)=3,\qquad f(1)=1.\]
Using Formula (1) the local linear approximation at $x_0=1$ is
\[L(x)=f(1)+f'(1)(x-1)=1+3(x-1).
\]
(b)} Writing $\Delta x=x-1$ (Formula (2) gives
\[L(x)=1+3\Delta x.
\]
{(c)} To approximate $(1.02)^3$ set $x=1.02$, so $\Delta x=0.02$.
\[L(1.02)=1+3(0.02)=1.06.
\]
For comparison, the exact value is
\[(1.02)^3=1.061208.\]
Thus, the linear approximation gives $1.06$, which differs from the exact value by
\[1.06-1.061208\approx -0.001208.\]
The result from part (b) is the same since it is algebraically identical.
