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Active Calculus 1st Ed

Preview Activity 8.5.1.
Preview Activity 8.3.1 showed how we can approximate the number \(e\) using linear, quadratic, and other polynomial functions; we then used similar ideas in Preview Activity 8.4.1 to approximate \(\ln(2)\text{.}\) In this activity, we review and extend the process to find the “best” quadratic approximation to the exponential function \(e^x\) around the origin. Let \(f(x) = e^x\) throughout this activity.
(a)
Find a formula for \(P_1(x)\text{,}\) the linearization of \(f(x)\) at \(x=0\text{.}\) (We label this linearization \(P_1\) because it is a first degree polynomial approximation.) Recall that \(P_1(x)\) is a good approximation to \(f(x)\) for values of \(x\) close to \(0\text{.}\) Plot \(f\) and \(P_1\) near \(x=0\) to illustrate this fact.
(b)
Since \(f(x) = e^x\) is not linear, the linear approximation eventually is not a very good one. To obtain better approximations, we want to develop a different approximation that “bends” to make it more closely fit the graph of \(f\) near \(x=0\text{.}\) To do so, we add a quadratic term to \(P_1(x)\text{.}\) In other words, we let
\begin{equation*} P_2(x) = P_1(x) + c_2x^2 \end{equation*}
for some real number \(c_2\text{.}\) We need to determine the value of \(c_2\) that makes the graph of \(P_2(x)\) best fit the graph of \(f(x)\) near \(x=0\text{.}\)
Remember that \(P_1(x)\) was a good linear approximation to \(f(x)\) near \(0\text{;}\) this is because \(P_1(0) = f(0)\) and \(P'_1(0) = f'(0)\text{.}\) It is therefore reasonable to seek a value of \(c_2\) so that
\begin{align*} P_2(0) \amp = f(0)\text{,} \amp P'_2(0) \amp = f'(0)\text{,} \amp \text{and }P''_2(0) \amp = f''(0)\text{.} \end{align*}
Remember, we are letting \(P_2(x) = P_1(x) + c_2x^2\text{.}\)
Calculate \(P_2(0)\) to show that \(P_2(0) = f(0)\text{.}\)
(c)
Calculate \(P'_2(0)\) to show that \(P'_2(0) = f'(0)\text{.}\)
(d)
Calculate \(P''_2(x)\text{.}\) Then find a value for \(c_2\) so that \(P''_2(0) = f''(0)\text{.}\)
(e)
Explain why the condition \(P''_2(0) = f''(0)\) will put an appropriate “bend” in the graph of \(P_2\) to make \(P_2\) fit the graph of \(f\) around \(x=0\text{.}\)