We are tasked with finding the values of $x$ where the function

$f(x) = \frac{e^{2x+4}}{2x}$

has a relative minimum. Let’s break this problem into steps:

Step 1: Find the derivative of $f(x)$

The function $f(x)$ is a quotient. Using the quotient rule, where $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, let: $u = e^{2x+4}, \quad v = 2x.$ We compute their derivatives: $u' = \frac{d}{dx}(e^{2x+4}) = 2e^{2x+4}, \quad v' = \frac{d}{dx}(2x) = 2.$

Now applying the quotient rule: $f'(x) = \frac{u'v - uv'}{v^2} = \frac{(2e^{2x+4})(2x) - (e^{2x+4})(2)}{(2x)^2}.$

Simplify the numerator: $f'(x) = \frac{4xe^{2x+4} - 2e^{2x+4}}{4x^2}.$ Factor $2e^{2x+4}$ from the numerator: $f'(x) = \frac{2e^{2x+4}(2x - 1)}{4x^2}.$ Simplify further: $f'(x) = \frac{e^{2x+4}(2x - 1)}{2x^2}.$

Step 2: Find critical points ($f'(x) = 0$ or undefined)

For $f'(x)$ to be zero, the numerator must equal zero: $e^{2x+4}(2x - 1) = 0.$ Since $e^{2x+4} > 0$ for all $x$, the equation reduces to: $2x - 1 = 0 \implies x = \frac{1}{2}.$

Next, check where $f'(x)$ is undefined. Since $f'(x)$ has a denominator $2x^2$, the derivative is undefined at $x = 0$. However, $f(x)$ is also undefined at $x = 0$ because the original function has $2x$ in the denominator. Thus, $x = 0$ is not a critical point.

Step 3: Determine if $x = \frac{1}{2}$ is a relative minimum

Use the first derivative test or the second derivative test.

First Derivative Test

Evaluate the sign of $f'(x)$ around $x = \frac{1}{2}$:

• For $x < \frac{1}{2}$, choose $x = 0.4$: $f'(0.4) = \frac{e^{2(0.4)+4}(2(0.4) - 1)}{2(0.4)^2} = \frac{e^{4.8}(-0.2)}{2(0.16)} < 0.$ Thus, $f'(x) < 0$ for $x < \frac{1}{2}$.

• For $x > \frac{1}{2}$, choose $x = 0.6$: $f'(0.6) = \frac{e^{2(0.6)+4}(2(0.6) - 1)}{2(0.6)^2} = \frac{e^{5.2}(0.2)}{2(0.36)} > 0.$ Thus, $f'(x) > 0$ for $x > \frac{1}{2}$.

Since $f'(x)$ changes from negative to positive at $x = \frac{1}{2}$, the function has a relative minimum at $x = \frac{1}{2}$.

Final Answer:

The function $f(x) = \frac{e^{2x+4}}{2x}$ has a relative minimum at: $x = \frac{1}{2}.$

Would you like more details on any step or another method of solving this problem?

Follow-up Questions:

1. How is the second derivative test applied to confirm a relative minimum?
2. Why does the factor $e^{2x+4}$ not affect critical points?
3. How does the domain of $f(x)$ influence its critical points?
4. What are other scenarios where the first derivative test is preferable?
5. Can the quotient rule be used on other types of problems like trigonometric functions?

Tip:

When solving problems involving exponential functions, remember that the exponential term is always positive, which simplifies finding critical points!