What is the geometric meaning of the degree-zero and degree-one Maclaurin approximations of $e^x$?

Conditions

• The function is $f(x) = e^x$.
• The approximations are centered at $x=0$.
• $S_0(x)$ is the constant term approximation.
• $S_1(x)$ is the linear term approximation.

Reasoning, step by step

1. Identify the degree-zero polynomial: $S_0(x) = 1$.
2. Observe that $S_0(0) = e^0 = 1$, matching the function value.
3. Identify the degree-one polynomial: $S_1(x) = 1 + x$.
4. Calculate the derivative of $S_1(x)$: $S_1'(x) = 1$.
5. Compare with $f'(x) = e^x$, so $f'(0) = 1$.
6. Conclude that $S_1(x)$ matches both value and slope at $x=0$, forming the tangent line.

Example

The script states: 'The constant approximation is $S_0(x)=1$. Adding x gives the tangent line $S_1(x)=1+x$.'

Common misconceptions

• Believing that $S_0(x)=1$ approximates the slope of the curve.
• Thinking that $S_1(x)$ matches the curvature of $e^x$ at $x=0$.

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