Exercise 1955Calculus › Single-variable calculus › Differentiation › Mean Value TheoremL1Added 4 Nov 2026

Second-order Taylor formula

#Taylor#MVT#second order

Statement

Let $f$ be $C^2$ on $[a,b]$ . Show that there exists $c\in]a,b[$ with
$f(b)=f(a)+f'(a)(b-a)+\frac{f''(c)}{2}(b-a)^2.$

Hint

Apply Rolle to $g(x)=f(b)-f(x)-f'(x)(b-x)-\frac{M}{2}(b-x)^2$ where $M$ is chosen so that $g(a)=0$ .

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Solution

Setup. Choose $M=\frac{2[f(b)-f(a)-f'(a)(b-a)]}{(b-a)^2}$ , so that $g(a)=0$ .

Rolle on $g$ . $g(b)=0$ , $g(a)=0$ ; there exists $c_1\in]a,b[$ with $g'(c_1)=0$ . Compute $g'(x)=-f''(x)(b-x)+M(b-x)=(b-x)[M-f''(x)]$ . Since $b-x>0$ on $]a,b[$ , $g'(c_1)=0$ gives $M=f''(c_1)$ .

Conclusion. $f(b)=f(a)+f'(a)(b-a)+\frac{f''(c_1)}{2}(b-a)^2$ .

Second-order Taylor formula

Statement

Solution

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