Let f: ℝ → ℝ and g: ℝ → ℝ be functions satisfying f(x + y) = f(x) + f(y) + f(x)f(y) and f(x) = xg(x)

Q: Wait, why did you add 1 to both sides of the functional equation? That seems like a random guess.

It is a classic technique in functional equations! The expression $f(x) + f(y) + f(x)f(y)$ is almost the expansion of $(1+f(x))(1+f(y))$. By adding 1 to both sides, we get $f(x+y) + 1 = f(x) + f(y) + f(x)f(y) + 1$, which factors beautifully into $f(x+y) + 1 = (1+f(x))(1+f(y))$. This transforms an additive-looking equation into a multiplicative one, which is much easier to handle.

Q: Why is $f'(0) = 1$?

We used the definition of the derivative at $x=0$. Since $f(0)=0$, the limit becomes $\lim_{h \to 0} \frac{f(h)}{h}$. Given $f(h) = hg(h)$, this simplifies to $\lim_{h \to 0} g(h)$, which is explicitly given as 1. It is a beautiful connection between the function's behavior at the origin and its slope.

Q: Is $g(x)$ really differentiable at $x=0$? It looks like it has a $0/0$ form.

Yes, it is! While $g(x) = \frac{e^x - 1}{x}$ is undefined at $x=0$ by simple substitution, the limit exists. To check differentiability, we calculate $\lim_{x \to 0} \frac{g(x) - g(0)}{x}$. Using the Taylor series $e^x = 1 + x + \frac{x^2}{2} + \dots$, the numerator becomes $\frac{x^2}{2} + \dots$, and dividing by $x^2$ leaves us with $1/2$. Since the limit is finite, it is indeed differentiable.

Question

Let $f: \mathbb{R} 	o \mathbb{R}$ and $g: \mathbb{R} 	o \mathbb{R}$ be functions satisfying $f(x + y) = f(x) + f(y) + f(x)f(y)$ and $f(x) = xg(x)$ for all $x, y \in \mathbb{R}$. If $\lim_{x 	o 0} g(x) = 1$, then which of the following statements is/are TRUE?

Options:

(A) (A) $f$ is differentiable at every $x \in \mathbb{R}$

(B) (B) If $g(0) = 1$, then $g$ is differentiable at every $x \in \mathbb{R}$

(C) (C) The derivative $f'(1)$ is equal to 1

(D) (D) The derivative $f'(0)$ is equal to 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are peeling back the layers of a beautiful functional equation.

We are given two functions, $f$ and $g$, linked by the relationship:
$$f(x + y) = f(x) + f(y) + f(x)f(y)$$

We are also given the definition $f(x) = xg(x)$ and the condition $\lim_{x 	o 0} g(x) = 1$. This is our compass. Let us begin.

### The Zero Insight

In any functional equation, the point $x=0$ is our anchor. Let us substitute $x=0$ into our relation $f(x) = xg(x)$.

The right side becomes $0 \cdot g(0)$, which is zero. Thus, we have our first solid ground:
$$f(0) = 0$$

This simple fact is the key that will unlock the derivative at the origin.

### The First Principle

Now, let us find $f'(0)$. We use the first principle of derivatives:
$$f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h}$$

Since $f(0) = 0$, this simplifies to $\lim_{h 	o 0} \frac{f(h)}{h}$. Substituting $f(h) = hg(h)$, we get:
$$\lim_{h 	o 0} \frac{hg(h)}{h} = \lim_{h 	o 0} g(h)$$

The problem explicitly told us this limit is $1$. Therefore, $f'(0) = 1$. This confirms that **Statement (D) is true**.

### The General Derivative

To find the derivative at any point $x$, we again turn to the first principle:
$$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$$

Here, we invoke the functional equation $f(x+h) = f(x) + f(h) + f(x)f(h)$. Substituting this into our limit, we get:
$$f'(x) = \lim_{h 	o 0} \frac{f(x) + f(h) + f(x)f(h) - f(x)}{h} = \lim_{h 	o 0} \frac{f(h)(1 + f(x))}{h}$$

Since $(1 + f(x))$ does not depend on $h$, we pull it out:
$$f'(x) = (1 + f(x)) \cdot \lim_{h 	o 0} \frac{f(h)}{h}$$

We already know $\lim_{h 	o 0} \frac{f(h)}{h} = f'(0) = 1$. Thus, $f'(x) = 1 + f(x)$. This derivative exists for all $x$, so **Statement (A) is true**.

### Solving the Differential Equation

We have arrived at a first-order differential equation: $f'(x) = 1 + f(x)$. This is a classic separable equation.

We write:
$$\frac{f'(x)}{1 + f(x)} = 1$$

Integrating both sides with respect to $x$, we obtain $\ln|1 + f(x)| = x + C$. Using our initial condition $f(0) = 0$, we find $C = 0$.

Exponentiating both sides, we get $1 + f(x) = e^x$, or:
$$f(x) = e^x - 1$$

With this, we can check Statement (C). Since $f'(1) = 1 + f(1) = 1 + (e^1 - 1) = e$, and $e 
eq 1$, **Statement (C) is false**.

### The Taylor Series Magic

Finally, let us examine $g(x)$. Since $f(x) = xg(x)$, for $x 
eq 0$, we have $g(x) = \frac{e^x - 1}{x}$.

To check if $g$ is differentiable at $x=0$, we calculate:
$$g'(0) = \lim_{x 	o 0} \frac{g(x) - g(0)}{x} = \lim_{x 	o 0} \frac{\frac{e^x - 1}{x} - 1}{x} = \lim_{x 	o 0} \frac{e^x - 1 - x}{x^2}$$

Using the Taylor series $e^x = 1 + x + \frac{x^2}{2!} + \dots$, the numerator becomes $\frac{x^2}{2} + \dots$. Dividing by $x^2$ gives $1/2$.

The limit exists, so $g$ is differentiable. **Statement (B) is true**. We have successfully navigated the problem!

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Visualized Solution

Analyzing the Given Functions & Finding $f (0)$

Evaluating $f^{'} (0)$

Derivative $f^{'} (x)$ via First Principles

Simplifying and Evaluating $f^{'} (x)$

Solving the Differential Equation for $f (x)$

Checking Statement (C)

Analyzing Function $g (x)$

Evaluating $g^{'} (0)$ with Taylor Series

The Sigma Insight: Differentiability of a Function

Analyzing the Setup

The Zero Insight

The First Principle

The General Derivative

Solving the Differential Equation

The Taylor Series Magic

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Visualized Solution

Analyzing the Given Functions & Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(0)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(0)

Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x) via First Principles

Simplifying and Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Solving the Differential Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)