Question

Let \(f:(-1,1)\to\mathbb{R}\) be a differentiable function. Consider the following statements:
(I) Suppose \(f(0)\geq0\), and \(f'(x)>0\) whenever \(f(x)=0\), for any \(x\geq0\). Then \(f(x)>0\), for any \(x>0\).
(II) Suppose \(f(0)\leq0\), and \(f'(x)>0\) whenever \(f(x)=0\), for any \(x\leq0\). Then \(f(x)<0\), for any \(x<0\).
Which of the following statements is correct?

• A. Only statement (I) is correct
• B. Only statement (II) is correct
• C. Both statements (I) and (II) are correct
• D. Neither statement (I) nor statement (II) is correct

Hint:

Use a contradiction argument: if f touches zero with a positive derivative there, it cannot fall back to zero or below right after, since that would force a non-positive slope at the crossing point. Reflect x to -x to turn statement (II) into statement (I).

Answer 1

The Correct Option is C

Step 1: Understand what needs to be shown.
Statement (I) says: if \(f(0)\geq0\) and the function has positive slope every time it touches zero for \(x\geq0\), then \(f\) stays strictly positive for all \(x>0\). Statement (II) is the mirror version on the left side, \(x\leq0\). We check each one by a "no return to zero" argument.

Step 2: Prove statement (I) when \(f(0)>0\).
Suppose, to get a contradiction, that \(f(x_1)\leq0\) for some \(x_1>0\). Since \(f\) is continuous (being differentiable) and \(f(0)>0\) while \(f(x_1)\leq0\), by the Intermediate Value Theorem there is a first point \(x_0\in(0,x_1]\) where \(f(x_0)=0\), with \(f(x)>0\) for all \(x\) in \([0,x_0)\).

Step 3: Derive the contradiction for (I).
Because \(f(x)>0\) just to the left of \(x_0\) and \(f(x_0)=0\), the left-hand difference quotient satisfies
\[ \frac{f(x_0)-f(x)}{x_0-x}\leq0 \quad\text{as } x\to x_0^{-}, \] so \(f'(x_0)\leq0\). But the hypothesis says \(f'(x)>0\) whenever \(f(x)=0\) for \(x\geq0\), and \(x_0\geq0\), \(f(x_0)=0\), so we need \(f'(x_0)>0\). This contradicts \(f'(x_0)\leq0\). So no such \(x_1\) exists, and \(f(x)>0\) for all \(x>0\).

Step 4: Handle the boundary case \(f(0)=0\) for (I).
If \(f(0)=0\), the hypothesis with \(x=0\) gives \(f'(0)>0\), so \(f\) is increasing through \(0\), meaning \(f(x)>0\) for all small \(x>0\). From that point the same "no return to zero" argument of Step 2 to Step 3 applies, so \(f(x)>0\) stays true for every \(x>0\). So statement (I) is TRUE in both cases.

Step 5: Prove statement (II) using a reflection trick.
Define \(g(s)=-f(-s)\) for \(s\geq0\). Then \(g(0)=-f(0)\geq0\), since \(f(0)\leq0\). Also \(g'(s)=f'(-s)\). Whenever \(g(s)=0\) we have \(f(-s)=0\), and since \(-s\leq0\), the hypothesis of (II) gives \(f'(-s)>0\), so \(g'(s)>0\). So \(g\) satisfies exactly the hypothesis of statement (I) (with \(g(0)\geq0\) and \(g'(s)>0\) whenever \(g(s)=0\), for \(s\geq0\)).

Step 6: Conclude statement (II).
By statement (I) applied to \(g\), we get \(g(s)>0\) for all \(s>0\), that is, \(-f(-s)>0\), so \(f(-s)<0\) for all \(s>0\). Writing \(x=-s<0\), this says \(f(x)<0\) for all \(x<0\), which is exactly statement (II). So (II) is also TRUE.

Final Answer:
Both statement (I) and statement (II) are correct. \[ \boxed{\text{Both statements (I) and (II) are correct}} \]