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?
Show 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).
Neither statement (I) nor statement (II) is correct
Show Solution
The Correct Option isC
Solution and Explanation
Step 1: Restate what each claim is really saying.
Both statements describe a function that cannot come back down through zero with a non-positive slope. Statement (I) looks at $x\geq0$ and rules out $f$ dropping to zero or below once it starts non-negative at $0$. Statement (II) is the same idea but read backwards from $0$ toward negative $x$.
Step 2: Set up statement (I) by contradiction.
Assume $f(0)\geq0$ and $f'(x)>0$ whenever $f(x)=0$ for $x\geq0$, but suppose $f(x_1)\leq0$ for some $x_1>0$. Continuity of $f$ (it is differentiable) plus the Intermediate Value Theorem gives a smallest crossing point $x_0\in(0,x_1]$ with $f(x_0)=0$ and $f>0$ just before $x_0$.
Step 3: Compare slopes at the crossing point.
Approaching $x_0$ from values where $f>0$ and hitting $f(x_0)=0$ means the function is falling into zero from above, so its derivative there cannot be positive: $f'(x_0)\leq0$. This clashes with the rule $f'(x_0)>0$ (since $f(x_0)=0$ and $x_0\geq0$). The contradiction kills the assumption, so $f$ stays positive for every $x>0$. If instead $f(0)=0$ itself, the same rule at $x=0$ forces $f'(0)>0$, pushing $f$ positive just after $0$, and the same crossing-point argument then keeps it positive forever after. So (I) holds.
Step 4: Prove (II) directly with the same style of argument, run leftward.
Assume $f(0)\leq0$ and $f'(x)>0$ whenever $f(x)=0$ for $x\leq0$. Suppose, for contradiction, $f(x_1)\geq0$ for some $x_1<0$. By the Intermediate Value Theorem there is a value $x_0\in[x_1,0)$ closest to $0$ with $f(x_0)=0$, and $f<0$ for all $x$ strictly between $x_0$ and $0$.
Step 5: Check the slope at that crossing.
Since $f<0$ immediately to the right of $x_0$ (between $x_0$ and $0$) and $f(x_0)=0$, the function rises from $0$ down into negative values as $x$ moves away from $x_0$. This means the right-hand difference quotient satisfies $\frac{f(x)-f(x_0)}{x-x_0}\leq0$ as $x\to x_0^{+}$, so $f'(x_0)\leq0$. But the hypothesis of (II) demands $f'(x_0)>0$ since $f(x_0)=0$ and $x_0\leq0$. This is a contradiction, so no such $x_1<0$ exists, meaning $f(x)<0$ for all $x<0$.
Step 6: Put both results together.
Both (I) and (II) survive their own contradiction test, so both statements are correct.
\[ \boxed{\text{Both statements (I) and (II) are correct}} \]
