Let f : R → (0, ∞) and g : R → R be twice differentiable functions such that f'' and g'' are continu

Q: Wait, why can we use L'Hopital's Rule here? Is it always applicable?

L'Hopital's Rule is a powerful tool, but it requires the limit to be in an indeterminate form like $0/0$ or $\infty/\infty$. Since we verified that $f(2)g(2) = 0$ and $f'(2)g'(2) = 0$, the conditions are perfectly met. It's not just a trick; it's a rigorous way to analyze the local behavior of functions near a point.

Q: How do we know for sure that $f(2) > 0$?

The problem defines the range of $f$ as $(0, \infty)$. This means the function $f$ is strictly positive for all real numbers $x$. Therefore, at any specific point, including $x=2$, $f(2)$ must be greater than zero. This is a vital piece of information that helps us determine the sign of the second derivative later.

Q: Why does $f''(2) = f(2)$ imply a local minimum?

The Second Derivative Test states that if $f'(c) = 0$ and $f''(c) > 0$, then $f$ has a local minimum at $c$. We found $f'(2) = 0$ (given) and $f''(2) = f(2)$. Since $f(2) > 0$, it follows that $f''(2) > 0$. Thus, the conditions for a local minimum are satisfied.

Question

Let $f : \mathbb{R} 	o (0, \infty)$ and $g : \mathbb{R} 	o \mathbb{R}$ be twice differentiable functions such that $f''$ and $g''$ are continuous functions on $\mathbb{R}$. Suppose $f'(2) = g(2) = 0, f''(2) 
eq 0$ and $g'(2) 
eq 0$. If $\lim_{x	o 2} \frac{f(x)g(x)}{f'(x)g'(x)} = 1$, then

Options:

(A) $f$ has a local minimum at $x = 2$

(B) $f$ has a local maximum at $x = 2$

(C) $f''(2) > f(2)$

(D) $f(x) - f''(x) = 0$ for at least one $x \in \mathbb{R}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, open field, and before you are two paths represented by the functions $f$ and $g$. The problem provides a beautiful constraint: the range of $f$ is $(0, \infty)$.

This means the graph of $f(x)$ is like a bird that never lands; it always hovers strictly above the $x$-axis. It can never touch the ground, and it can never dive below it.

Meanwhile, $g(x)$ is a more adventurous path, crossing the $x$-axis at $x=2$. This is our point of interest, the coordinate where the magic happens.

### The Indeterminate Mystery

We are presented with a limit:

$$\lim_{x	o 2} \frac{f(x)g(x)}{f'(x)g'(x)} = 1$$

Before we panic, let's test the waters. What happens at $x=2$? The numerator becomes $f(2)g(2)$. Since $g(2)=0$, the numerator is zero.

The denominator is $f'(2)g'(2)$. We are told $f'(2)=0$, so the denominator is also zero. We are staring at a $0/0$ indeterminate form.

This is not a dead end; it is an invitation to use L'Hopital's Rule. We are essentially looking for the ratio of the rates of change of these functions as they approach the point $x=2$.

### The Calculus of Change

To apply L'Hopital's Rule, we must differentiate the numerator and the denominator. This is where the Product Rule becomes our best friend.

The derivative of the numerator $f(x)g(x)$ is $f'(x)g(x) + f(x)g'(x)$. Similarly, the derivative of the denominator $f'(x)g'(x)$ is $f''(x)g'(x) + f'(x)g''(x)$.

Our limit now looks like this:

$$\lim_{x	o 2} \frac{f'(x)g(x) + f(x)g'(x)}{f''(x)g'(x) + f'(x)g''(x)} = 1$$

This looks intimidating, but let's breathe and substitute $x=2$. We know $f'(2)=0$ and $g(2)=0$.

The numerator simplifies beautifully: $0 \cdot 0 + f(2)g'(2) = f(2)g'(2)$. The denominator simplifies to $f''(2)g'(2) + 0 \cdot g''(2) = f''(2)g'(2)$.

### The Elegant Cancellation

Now we have:

$$\frac{f(2)g'(2)}{f''(2)g'(2)} = 1$$

The problem explicitly states $g'(2) 
eq 0$, which is our golden ticket to cancel $g'(2)$ from both the numerator and the denominator.

We are left with the elegant relationship:

$$\frac{f(2)}{f''(2)} = 1 \quad \Rightarrow \quad f''(2) = f(2)$$

This is the key that unlocks the entire problem.

### The Final Revelation

We established that $f(x) > 0$ for all $x$, so $f(2) > 0$. Since $f''(2) = f(2)$, it must be that $f''(2) > 0$.

We have a point $x=2$ where the first derivative $f'(2)$ is zero (a horizontal tangent) and the second derivative $f''(2)$ is positive (concave up). By the Second Derivative Test, this point is a **local minimum**.

Furthermore, since $f''(2) = f(2)$, the equation $f(x) - f''(x) = 0$ is satisfied at $x=2$. We have successfully navigated the landscape of this problem and uncovered its hidden truths.

Select Answer:

Visualized Solution

Understanding the Functions

Conditions at $x = 2$

Analyzing the Limit

The Indeterminate Form

Applying L'Hopital's Rule

Using the Product Rule

Substituting $x = 2$ Again

Simplifying the Expression

Canceling $g^{'} (2)$

Analyzing the Sign of $f^{''} (2)$

The Second Derivative Test

Checking the Remaining Options

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Indeterminate Mystery

The Calculus of Change

The Elegant Cancellation

The Final Revelation

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