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*}
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