Why does the video use value, slope, and concavity to determine a quadratic approximation to $e^x$ at $x=0$?

Conditions

• The approximating polynomial is quadratic.
• The matching point is $x=0$.
• The target function and polynomial can be differentiated at least twice at that point.

Reasoning, step by step

1. Start with a generic quadratic approximation to $e^x$ near $x=0$: $c_0 + c_1x + c_2x^2$.
2. Impose the same-value condition at $x=0$ to determine $c_0$.
3. Impose the same-slope condition at $x=0$ by differentiating both sides and evaluating at $x=0$ to determine $c_1$.
4. Impose the same-concavity condition at $x=0$ by taking the second derivative and evaluating at $x=0$ to determine $c_2$.
5. Substitute the solved coefficients back into the generic quadratic to obtain the final approximation.

Example

The speaker states that the quadratic has three different restrictions and a generic quadratic has three different coefficients. The board displays the generic quadratic $c_0+c_1x+c_2x^2$ and the graph labels the contact point with “Same value,” “Same slope,” and then “Same concavity.”

Common misconceptions

• Believing that matching only the value and slope is sufficient for a quadratic approximation, leaving the curvature unconstrained.
• Assuming that the three conditions prove a general theorem of uniqueness for all functions, rather than serving as the motivation for the worked setup.

Watch the explanation

Explore next

• How are $c_0$, $c_1$, and $c_2$ found for the approximation of $e^x$ at $x=0$?
• What does “same concavity” mean mathematically in this Taylor-series setup?