Strict Monotonicity and a Zero Derivative

Detailed Solution

We are given that $f$ is differentiable on $\mathbb R$ and $$ f'(x)>0\qquad\text{for every }x\ne0, $$ while $$ f'(0)=0. $$ We will show that $f$ is strictly increasing on $\mathbb R$. **Step 1: Establish that $f$ is non-decreasing.** Since $f$ is differentiable, it is continuous on $\mathbb R$. For any $a,b\in\mathbb R$ with $a<b$, the Mean Value Theorem guarantees the existence of $c\in(a,b)$ such that $$ f(b)-f(a)=f'(c)(b-a). $$ Because $f'(c)\ge0$ for every $c\in\mathbb R$, we obtain $$ f(b)-f(a)\ge0. $$ Therefore, $$ f(a)\le f(b), $$ so $f$ is non-decreasing. **Step 2: Establish strict increase.** Let $a<b$. We again apply the Mean Value Theorem: $$ f(b)-f(a)=f'(c)(b-a) $$ for some $c\in(a,b)$. If $c\ne0$, then $f'(c)>0$, and hence $$ f(b)-f(a)>0. $$ If $c=0$, this particular application does not establish strict inequality. However, because $f'(x)>0$ at every point other than $0$, the function is strictly increasing on both $(-\infty,0)$ and $(0,\infty)$. Moreover, differentiability at $0$ and the positivity of the derivative on either side prevent any decrease across the origin. Consequently, $$ f(a)<f(b)\qquad\text{whenever }a<b. $$ Thus $f$ is strictly increasing on $\mathbb R$. **Verification of the options** - **A is true.** A differentiable function whose derivative is positive everywhere except at one point where it vanishes is still strictly increasing. - **B is false.** Strict increase means that $f$ cannot have a local minimum at $0$ in the usual two-sided sense unless the function's values on either side are considered; in fact, a strictly increasing function does not have a local minimum at an interior point. - **C is false.** The function is strictly increasing, not merely non-decreasing. - **D is false.** The condition $f'(0)=0$ does not force an inflection point. For example, $f(x)=x^3$ has a stationary inflection point at $0$, but $f(x)=x+x^3$ has $f'(0)=1$, so the vanishing derivative alone does not determine the concavity behavior. Therefore, the correct option is $$ \boxed{\mathrm{A}}. $$