Second Derivative Test at a Local Maximum

Detailed Solution

We are given that $f$ is differentiable on $\mathbb R$, that $f'$ is continuous, and that $f''(x_0)$ exists. Also, $f(x_0)$ is a local maximum. We will show that $$ f'(x_0)=0 \quad\text{and}\quad f''(x_0)\le 0. $$ **Step 1: Prove that $f'(x_0)=0$.** Since $f(x_0)$ is a local maximum, there exists $\delta>0$ such that $$ f(x)\le f(x_0) $$ whenever $|x-x_0|<\delta$. For $h>0$ sufficiently small, $$ \frac{f(x_0+h)-f(x_0)}{h}\le0. $$ Taking the right-hand limit gives $$ f'_+(x_0)\le0. $$ For $h<0$ sufficiently small, division by $h$ reverses the inequality: $$ \frac{f(x_0+h)-f(x_0)}{h}\ge0. $$ Taking the left-hand limit gives $$ f'_-(x_0)\ge0. $$ Because $f$ is differentiable at $x_0$, both one-sided derivatives are equal. Hence, $$ \boxed{f'(x_0)=0}. $$ **Step 2: Prove that $f''(x_0)\le0$.** Since $f''(x_0)$ exists and $f'(x_0)=0$, we have $$ f''(x_0) =\lim_{h\to0}\frac{f'(x_0+h)-f'(x_0)}{h}. $$ Equivalently, $$ f''(x_0) =\lim_{h\to0}\frac{f'(x_0+h)}{h}. $$ A more direct argument uses the second-order Taylor expansion: $$ f(x_0+h) =f(x_0)+f'(x_0)h+\frac12f''(x_0)h^2+o(h^2). $$ Since $f'(x_0)=0$, $$ f(x_0+h)-f(x_0) =\frac12f''(x_0)h^2+o(h^2). $$ At a local maximum, the left-hand side is non-positive for sufficiently small $h$. Dividing by $h^2>0$ and taking the limit yields $$ \frac12f''(x_0)\le0. $$ Therefore, $$ \boxed{f''(x_0)\le0}. $$ **Step 3: Explain why strict negativity is not necessary.** Consider the function $$ f(x)=-x^4. $$ At $x_0=0$, the function has a strict local maximum because $$ f(x)<f(0)=0\qquad(x\ne0). $$ However, $$ f'(x)=-4x^3,\qquad f''(x)=-12x^2. $$ Thus, $$ f'(0)=0,\qquad f''(0)=0. $$ So a local maximum can occur even when the second derivative is zero. **Verification of the options** - **A is false:** A local maximum does not require $f''(x_0)<0$. - **B is true:** We must have $f'(x_0)=0$ and $f''(x_0)\le0$. - **C is true as a sufficient condition** when $f''(x_0)<0$ and $f'$ is continuous near $x_0$, but it is not the necessary conclusion requested. - **D is false:** The example $f(x)=-x^4$ has a strict local maximum at $0$ with both derivatives equal to zero. Therefore, the intended correct answer is $$ \boxed{\mathrm{B}}. $$