Exercise 1.
\(x_0=10\text{,}\) \(f(10)=50\text{,}\) \(mf(11)=6\text{,}\) and \(x_1=15\text{.}\)
Solution.
\begin{align*}
f(x) \amp \approx f(10)+mf(11)*(x-10)\\
f(x) \amp \approx 50+6*(x-10)\text{.}
\end{align*}
Next we evaluate the linear approximation at 15:
\begin{equation*}
f(15)\approx 50+6*(15-10)=50+30=80\text{.}
\end{equation*}