Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be such that $f(2+x)=f(2-x)\,\,\forall x\in\mathbb{R}$. If $f(x)$ is twice differentiable such that $f^{\prime}(1)=0$, then which one of the following is true?

• A. there exist at least one $c$ in $(0, 1)$ such that $f^{\prime}(c)=0$
• B. there exist at least one $c in (1, 2)$ such that $f^{\prime\prime}(c)=0$
• C. there exist at least one $c$ in $(0, 1)$ such that $f^{\prime\prime}(c)=0$
• D. there exist at least one $c$ in $(1, 2)$ such that $f^{\prime}(c)=0$

Hint:

Differentiating functional symmetry equations like $f(a+x)=f(a-x)$ immediately provides helpful values for the derivative function $f^{\prime}(x)$.

Answer 1

The Correct Option is B

Step 1: Read the symmetry.
The rule $f(2+x)=f(2-x)$ means the graph is a mirror image about the vertical line $x=2$.
Step 2: Differentiate the symmetry.
Differentiating both sides with respect to $x$ gives $f'(2+x)=-f'(2-x)$.
Step 3: Find a hidden zero of $f'$.
Put $x=0$: $f'(2)=-f'(2)$, so $2f'(2)=0$, meaning $f'(2)=0$. This is always true from the symmetry.
Step 4: Use the given zero.
We are also told $f'(1)=0$. So now we know $f'(1)=0$ and $f'(2)=0$, both inside the same interval $[1,2]$.
Step 5: Apply Rolle's theorem to $f'$.
Since $f$ is twice differentiable, $f'$ is continuous and differentiable. With $f'(1)=f'(2)=0$, Rolle's theorem on $f'$ gives a point $c\in(1,2)$ where $(f')'(c)=0$, that is $f''(c)=0$.
Step 6: Pick the correct statement.
So there is at least one $c\in(1,2)$ with $f''(c)=0$. (Note $f'(c)=0$ inside $(1,2)$ is not guaranteed.) \[ \boxed{\exists\,c\in(1,2):\ f''(c)=0} \]