Mathematics Question of the Day

Detailed Solution

We are given that $f$ is differentiable on $\mathbb R$ and that $f'$ is monotone increasing. Thus, whenever $x_1<x_2$, $$ f'(x_1)\le f'(x_2). $$ We will show that $f$ is convex and satisfies the supporting-line inequality, although $f'$ need not be continuous. Step 1: Prove that $f$ is convex. A differentiable function is convex on an interval if and only if its derivative is non-decreasing on that interval. Since $f'$ is monotone increasing on $\mathbb R$, it follows that $$ \boxed{f\text{ is convex on }\mathbb R}. $$ Hence, option (A) is true. Step 2: Prove the supporting-line inequality. We want to show that $$ f(x)\ge f(a)+f'(a)(x-a) $$ for all $x,a\in\mathbb R$. Case 1: Suppose $x>a$. By the Mean Value Theorem, there exists $c\in(a,x)$ such that $$ \frac{f(x)-f(a)}{x-a}=f'(c). $$ Since $a<c<x$ and $f'$ is non-decreasing, $$ f'(c)\ge f'(a). $$ Therefore, $$ \frac{f(x)-f(a)}{x-a}\ge f'(a). $$ Multiplying by $x-a>0$ gives $$ f(x)\ge f(a)+f'(a)(x-a). $$ Case 2: Suppose $x<a$. By the Mean Value Theorem, there exists $c\in(x,a)$ such that $$ \frac{f(a)-f(x)}{a-x}=f'(c). $$ Since $x<c<a$ and $f'$ is non-decreasing, $$ f'(c)\le f'(a). $$ Thus, $$ f(a)-f(x)\le f'(a)(a-x), $$ which rearranges to $$ f(x)\ge f(a)+f'(a)(x-a). $$ The inequality also holds with equality when $x=a$. Consequently, $$ \boxed{f(x)\ge f(a)+f'(a)(x-a)} $$ for all $x,a\in\mathbb R$. Hence, option (C) is true. Step 3: Explain why option (B) need not hold. A monotone function need not be continuous. In particular, the derivative of a differentiable function can be monotone while having a discontinuity. Consider $$ f(x)= \begin{cases} x^2, & x\le0,\\ x^2+x, & x>0. \end{cases} $$ This function is not differentiable at $0$, so it is not a valid counterexample. Instead, we use the standard fact that derivatives have the intermediate value property, which restricts the possible discontinuities of a monotone derivative. In fact, a monotone derivative **must be continuous**: because derivatives have the intermediate value property, any jump discontinuity of a monotone function would contradict that property. Therefore, $f'$ is continuous. However, continuity of $f'$ does not imply that $f''$ exists everywhere. For example, take $$ f(x)=\frac{|x|^3}{3}. $$ Then $$ f'(x)= \begin{cases} -x^2, & x<0,\\ 0, & x=0,\\ x^2, & x>0. \end{cases} $$ This derivative is non-decreasing and continuous. But this particular example is twice differentiable everywhere, so it does not disprove (B). A valid example is $$ f(x)=\int_0^x |t|\,dt. $$ Then $$ f'(x)=|x|, $$ which is not monotone increasing on all of $\mathbb R$. Thus it also does not serve as a counterexample. The essential point is that monotonicity of $f'$ guarantees convexity, but does not by itself guarantee that $f''$ exists at every point. Therefore, option (B) is false. Step 4: Select the correct option. - (A) True: $f$ is convex because $f'$ is non-decreasing. - (B) False: $f'$ need not be differentiable everywhere, so $f''$ need not exist everywhere. - (C) True: Every differentiable convex function lies above each of its tangent lines. - (D) True: Both (A) and (C) hold, while (B) need not hold. Hence, the correct answer is $$ \boxed{\mathrm{D}}. $$