Question

For all twice differentiable functions \(f:\mathbb{R}\to\mathbb{R}\), with \(f(0)=f(1)=f'(0)=0\)

• A. \(f''(x)\ne 0\), at every point \(x\in(0,1)\)
• B. \(f''(x)=0\), for some \(x\in(0,1)\)
• C. \(f''(0)=0\)
• D. \(f''(x)=0\), at every point \(x\in(0,1)\)

Hint:

Apply Rolle's theorem to f on [0,1], then again to f' on a subinterval.

Answer 1

The Correct Option is B

Since \(f(0)=f(1)=0\) and f is differentiable, Rolle's theorem gives some \(c\in(0,1)\) with \(f'(c)=0\).
Now \(f'(0)=0=f'(c)\) and \(f'\) is differentiable (f is twice differentiable), so Rolle's theorem applied to \(f'\) on \([0,c]\) gives some \(d\in(0,c)\subset(0,1)\) with \(f''(d)=0\).
Thus \(f''(x)=0\) for some \(x\in(0,1)\). The other options fail, e.g. for \(f(x)=x^2(x-1)\) we have \(f''(x)=6x-2\), which is non-zero at most points and nonzero at 0.
Answer: (2) \(f''(x)=0\), for some \(x\in(0,1)\)