Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
d. Suppose f'' is continuous on an interval that contains a, where f has an inflection point at a. Then the second−order Taylor polynomial for f at a is linear.

Verified step by step guidance

1. Recall that an inflection point at \(x = a\) means the concavity of the function changes at \(a\), which implies that the second derivative \(f''(a) = 0\) if \(f''\) is continuous around \(a\).
2. The second-order Taylor polynomial for \(f\) at \(a\) is given by: \[T_2(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^2\]
3. Since \(f''(a) = 0\) at the inflection point, the quadratic term in the Taylor polynomial disappears, leaving: \[T_2(x) = f(a) + f'(a)(x - a)\]
4. This polynomial is linear in \((x - a)\) because it only contains terms up to the first power of \((x - a)\).
5. Therefore, the second-order Taylor polynomial at an inflection point where \(f''\) is continuous is indeed linear, confirming the statement.

Key Concepts

Inflection Point

An inflection point of a function f is a point where the concavity changes, meaning f'' changes sign. At this point, the second derivative f''(a) is typically zero or undefined, indicating a transition between concave up and concave down behavior.

Second-Order Taylor Polynomial

The second-order Taylor polynomial of a function f at a point a approximates f near a using terms up to the second derivative: P_2(x) = f(a) + f'(a)(x - a) + (f''(a)/2)(x - a)^2. Its degree depends on whether f''(a) is zero or not.

Continuity of the Second Derivative

If f'' is continuous near a, then f''(a) exists and is finite. For an inflection point at a with continuous f'', f''(a) = 0, which affects the form of the Taylor polynomial by eliminating the quadratic term, making it linear.