Mean Value Theorem and Multiple Secant Slopes

Detailed Solution

We are given that $f$ is differentiable on $\mathbb R$ and $$ f(0)=0,\qquad f(1)=1,\qquad f(2)=1. $$ We will apply the Mean Value Theorem on two different intervals. Step 1: Prove that option (A) must hold. Consider the interval $[0,2]$. Since $f$ is differentiable on $\mathbb R$, it is continuous on $[0,2]$ and differentiable on $(0,2)$. By the Mean Value Theorem, there exists $c_1\in(0,2)$ such that $$ f'(c_1)=\frac{f(2)-f(0)}{2-0}. $$ Substituting the given values, $$ f'(c_1)=\frac{1-0}{2}=\frac12. $$ Therefore, $$ \boxed{\exists\,c_1\in(0,2):f'(c_1)=\frac12}. $$ Hence, option (A) is true. Step 2: Prove that option (B) must hold. Consider the interval $[1,2]$. Again, the Mean Value Theorem guarantees the existence of $c_2\in(1,2)$ such that $$ f'(c_2)=\frac{f(2)-f(1)}{2-1}. $$ Using $f(1)=f(2)=1$, we obtain $$ f'(c_2)=\frac{1-1}{1}=0. $$ Since $(1,2)\subset(0,2)$, we have $$ \boxed{\exists\,c_2\in(0,2):f'(c_2)=0}. $$ Therefore, option (B) is also true. Step 3: Determine the correct option. We have established both results: $$ f'(c_1)=\frac12 \quad\text{for some }c_1\in(0,2), $$ and $$ f'(c_2)=0 \quad\text{for some }c_2\in(1,2). $$ Consequently, both (A) and (B) must hold simultaneously. Verification of the options: (A) True. The Mean Value Theorem on $[0,2]$ guarantees a point where $f'(c)=\frac12$. (B) True. The Mean Value Theorem on $[1,2]$ guarantees a point where $f'(c)=0$. (C) True. Both (A) and (B) are necessarily satisfied. (D) False. The Mean Value Theorem on $[1,2]$ guarantees a point where $f'(c)=0$, not necessarily one where $f'(c)=1$. Therefore, the correct answer is $$ \boxed{\mathrm{C}}. $$