Let g(x) be a linear function and f(x) = {g(x) , x ≤ 0; ((1+x)/(2+x))^(1/x) , x 0}, is continuous

Q: Wait, why is the limit of $(\frac{1+x}{2+x})^{1/x}$ equal to $0$ as $x \to 0^+$?

Great question! Look at the base: as $x \to 0^+$, the fraction $\frac{1+x}{2+x}$ approaches $\frac{1+0}{2+0} = \frac{1}{2}$. Now look at the exponent: $\frac{1}{x}$ as $x \to 0^+$ approaches $+\infty$. So, you are essentially calculating $(\frac{1}{2})^\infty$. Since $\frac{1}{2}$ is a fraction less than $1$, raising it to an infinitely large power causes it to shrink to $0$.

Q: Why do we use logarithmic differentiation for the right branch?

Standard power rules like $\frac{d}{dx}(x^n) = nx^{n-1}$ only work when the exponent is a constant. Here, both the base and the exponent depend on $x$. Logarithmic differentiation allows us to bring that exponent down using the property $\ln(u^v) = v \ln(u)$, transforming the complex exponential into a product that is much easier to differentiate using the product rule.

Q: How do I know $g(x) = ax$ and not $ax+b$?

We determined $b$ from the continuity condition at $x=0$. Since $f(x)$ is continuous at $x=0$, the left-hand limit must equal the right-hand limit. We found the right-hand limit to be $0$, and the left-hand limit is $g(0) = a(0) + b = b$. Therefore, $b$ must be $0$, leaving us with $g(x) = ax$.

Question

Let $g(x)$ be a linear function and $f(x) = \begin{cases} g(x) & , x \le 0 \ (\frac{1+x}{2+x})^{1/x} & , x > 0 \end{cases}$, is continuous at $x = 0$. If $f'(1) = f(-1)$, then the value of $g(3)$ is

Options:

(A) $\frac{1}{3} \log_e(\frac{4}{9e^{1/3}})$

(B) $\frac{1}{3} \log_e(\frac{4}{9}) + 1$

(C) $\log_e(\frac{4}{9}) - 1$

(D) $\log_e(\frac{4}{9e^{1/3}})$

Sigma Percentile · Accepted Answer

### Analyzing the Bridge of Continuity

We are given a piecewise function $f(x)$. For $x \le 0$, it is defined by the linear function $g(x) = ax + b$. For $x > 0$, it is defined as $f(x) = \left( \frac{1+x}{2+x} ight)^{1/x}$.

For this function to be continuous at $x=0$, the two paths must meet at the same height. Mathematically, this requires:
$$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$$

Evaluating the left side is straightforward: $f(0) = g(0) = b$. Now, we evaluate the right-hand limit:
$$\lim_{x 	o 0^+} \left( \frac{1+x}{2+x} ight)^{1/x}$$

As $x$ approaches $0$ from the positive side, the base $\frac{1+x}{2+x}$ approaches $\frac{1}{2}$. Meanwhile, the exponent $\frac{1}{x}$ approaches infinity. Since $\left( \frac{1}{2} ight)^\infty = 0$, we conclude that $b=0$. Thus, our linear function simplifies to $g(x) = ax$.

### The Surgical Precision of Logarithmic Differentiation

We must satisfy the condition $f'(1) = f(-1)$. Since $f(x) = \left( \frac{1+x}{2+x} ight)^{1/x}$ involves a variable in both the base and the exponent, we apply logarithmic differentiation.

Taking the natural logarithm of both sides:
$$\ln f(x) = \frac{1}{x} \ln \left( \frac{1+x}{2+x} ight) = \frac{1}{x} [\ln(1+x) - \ln(2+x)]$$

Differentiating both sides with respect to $x$ using the product rule:
$$\frac{f'(x)}{f(x)} = -\frac{1}{x^2} \ln \left( \frac{1+x}{2+x} ight) + \frac{1}{x} \left[ \frac{1}{1+x} - \frac{1}{2+x} ight]$$

### Evaluating the Derivative

To find $f'(1)$, we first note that $f(1) = \left( \frac{1+1}{2+1} ight)^1 = \frac{2}{3}$. Substituting $x=1$ into the derivative expression:
$$f'(1) = \frac{2}{3} \left[ -\ln \left( \frac{2}{3} ight) + \left( \frac{1}{2} - \frac{1}{3} ight) ight]$$

Simplifying the terms inside the bracket:
$$f'(1) = \frac{2}{3} \left[ -\ln \left( \frac{2}{3} ight) + \frac{1}{6} ight]$$

### The Grand Synthesis

We know $f(-1) = g(-1) = a(-1) = -a$. Given the condition $f'(1) = f(-1)$, we equate the expressions:
$$\frac{2}{3} \left[ -\ln \left( \frac{2}{3} ight) + \frac{1}{6} ight] = -a$$

Distributing the $\frac{2}{3}$ and solving for $a$:
$$-\frac{2}{3} \ln \left( \frac{2}{3} ight) + \frac{1}{9} = -a \implies a = \frac{2}{3} \ln \left( \frac{2}{3} ight) - \frac{1}{9}$$

Finally, we calculate $g(3) = 3a$:
$$g(3) = 3 \left[ \frac{2}{3} \ln \left( \frac{2}{3} ight) - \frac{1}{9} ight] = 2 \ln \left( \frac{2}{3} ight) - \frac{1}{3}$$

Using logarithmic properties, $2 \ln \left( \frac{2}{3} ight) = \ln \left( \frac{4}{9} ight)$ and $\frac{1}{3} = \ln(e^{1/3})$. The final result is:
**$$g(3) = \ln \left( \frac{4}{9e^{1/3}} ight)$$**

Animated Solution for Mathematics - Differentiation: Let $g (x)$ be a linear function and $f (x) = {g (x) (\frac{1 + x}{2 + x})^{1/ x}, x \leq 0, x > 0$ , is continuous at $x = 0$ . If $f^{'} (1) = f (- 1)$ , then the value of $g (3)$ is

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Visualized Solution

Defining the Linear Function $g (x)$

Condition for Continuity at $x = 0$

Evaluating the Limit as $x \to 0^{+}$

Simplifying $g (x)$ and Finding $f (- 1)$

Logarithmic Differentiation for $x > 0$

Differentiating with respect to $x$

Evaluating $f (1)$ and $f^{'} (1)$

Solving for the Constant $a$

Calculating the Final Value $g (3)$

Final Simplification

The Sigma Insight: Techniques of Differentiation

Analyzing the Bridge of Continuity

The Surgical Precision of Logarithmic Differentiation

Evaluating the Derivative

The Grand Synthesis

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Visualized Solution

Defining the Linear Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Condition for Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Evaluating the Limit as .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x→0+

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

Logarithmic Differentiation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>0

Differentiating with respect to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

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

Solving for the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Calculating the Final Value .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(3)