Let α be a non-zero real number. Suppose f : R → R is a differentiable function such that f(0) = 2 a

Q: Wait, why can't I just use separation of variables here? It looks like a simple equation.

That is a very common trap! Separation of variables requires the equation to be in the form $f'(x) = g(x)h(f(x))$. Because of the additive constant $+3$, you cannot isolate the $f(x)$ terms and the $dx$ terms on opposite sides. This structure demands the linear differential equation method.

Q: How do I know if $\alpha$ should be positive or negative? I got stuck at the limit step.

Great question! Look at the term $C e^{\alpha x}$. If $\alpha$ were positive, then as $x \to -\infty$, $e^{\alpha x}$ would approach $0$, but if we were looking at $x \to \infty$, it would explode. More importantly, the limit condition $\lim_{x \to -\infty} f(x) = 1$ forces the exponential term to vanish or stabilize. By setting the constant term $-\frac{3}{\alpha}$ to $1$, we mathematically force $\alpha$ to be $-3$, which ensures the function converges to $1$ as $x$ goes to negative infinity.

Q: Is it always true that $e^{\ln 8} = 8$? I always get nervous with log identities.

Yes, absolutely! Remember that the natural logarithm $\ln$ and the exponential function $e^x$ are inverse functions. They essentially 'undo' each other. Whenever you see $e^{\ln(\text{anything})}$, you can confidently simplify it to just that 'anything'. It is one of the most powerful shortcuts in your toolkit!

Question

Let $\alpha$ be a non-zero real number. Suppose $f : R 	o R$ is a differentiable function such that $f(0) = 2$ and $\lim_{x	o-\infty} f(x) = 1$. If $f'(x) = \alpha f(x) + 3$, for all $x \in R$, then $f(-\ln_e 2)$ is equal to

Options:

(A) 3

(B) 5

(C) 9

(D) 7

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The differential equation is given by $f'(x) = \alpha f(x) + 3$. This represents a system striving for equilibrium, governed by the laws of growth and decay.

We rewrite the equation in the standard form:
$$f'(x) - \alpha f(x) = 3$$

This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -\alpha$ and $Q(x) = 3$.

### The Integrating Factor

To solve this, we utilize the Integrating Factor ($IF$), defined as $IF = e^{\int P(x) dx}$. Substituting our value for $P(x)$, we obtain:
$$IF = e^{\int -\alpha dx} = e^{-\alpha x}$$

Multiplying the entire differential equation by $e^{-\alpha x}$ allows us to express the left side as the derivative of a product:
$$\frac{d}{dx} [f(x) e^{-\alpha x}] = 3 e^{-\alpha x}$$

### The Integration

Integrating both sides with respect to $x$ yields:
$$\int \frac{d}{dx} [f(x) e^{-\alpha x}] dx = \int 3 e^{-\alpha x} dx$$

Performing the integration, we get:
$$f(x) e^{-\alpha x} = -\frac{3}{\alpha} e^{-\alpha x} + C$$

Multiplying through by $e^{\alpha x}$ isolates the general solution:
$$f(x) = -\frac{3}{\alpha} + C e^{\alpha x}$$

### The Detective Work

We are given the boundary condition $\lim_{x 	o -\infty} f(x) = 1$. For this limit to exist as a finite value, the exponential term must vanish as $x 	o -\infty$, which requires $\alpha < 0$.

Under this condition, the limit simplifies to the constant term:
$$-\frac{3}{\alpha} = 1 \Rightarrow \alpha = -3$$

### Pinning Down the Curve

Substituting $\alpha = -3$ into our general solution, we obtain $f(x) = 1 + C e^{-3x}$. We apply the initial condition $f(0) = 2$ to find the constant $C$:
$$2 = 1 + C e^{0} \Rightarrow 2 = 1 + C \Rightarrow C = 1$$

Thus, the specific function describing our system is:
$$f(x) = 1 + e^{-3x}$$

### Final Calculation

We evaluate the function at $x = -\ln 2$:
$$f(-\ln 2) = 1 + e^{-3(-\ln 2)}$$

Simplifying the exponent:
$$f(-\ln 2) = 1 + e^{3 \ln 2} = 1 + e^{\ln(2^3)} = 1 + e^{\ln 8}$$

Since $e^{\ln 8} = 8$, the final result is:
**$f(-\ln 2) = 9$**

Animated Solution for Mathematics - Differential Equations: Let $α$ be a non-zero real number. Suppose $f : R \to R$ is a differentiable function such that $f (0) = 2$ and $lim_{x \to - \infty} f (x) = 1$ . If $f^{'} (x) = α f (x) + 3$ , for all $x \in R$ , then $f (- ln_{e} 2)$ is equal to

Select Answer:

Visualized Solution

Standard Linear Form

Integrating Factor ( $I F$ )

General Solution Setup

Integration and Isolation

Applying the Limit

Updated Function

Initial Condition

Final Evaluation

Logarithmic Simplification

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Integrating Factor

The Integration

The Detective Work

Pinning Down the Curve

Final Calculation

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