We are tasked with using the first derivative test to find the location and value of any local extrema of the function:

$f(x) = x e^{5x} - 3$

Step 1: Compute the first derivative of $f(x)$

To apply the first derivative test, we first need to find the first derivative $f'(x)$. We use the product rule and the chain rule to differentiate:

$f'(x) = \frac{d}{dx}(x e^{5x}) - 0 \quad \text{(since the derivative of -3 is 0)}$

Now apply the product rule:

$f'(x) = \frac{d}{dx}(x) \cdot e^{5x} + x \cdot \frac{d}{dx}(e^{5x})$

$f'(x) = e^{5x} + x \cdot 5e^{5x}$

So, the first derivative is:

$f'(x) = e^{5x}(1 + 5x)$

Step 2: Set $f'(x) = 0$ to find critical points

To find the critical points where the function could have local extrema, we set $f'(x) = 0$:

$e^{5x}(1 + 5x) = 0$

Since $e^{5x}$ is never zero for any real value of $x$, the equation simplifies to:

$1 + 5x = 0$

Solving for $x$:

$x = -\frac{1}{5}$

Thus, we have a single critical point at $x = -\frac{1}{5}$.

Step 3: Determine the nature of the critical point using the first derivative test

To determine whether this critical point is a local minimum, local maximum, or a saddle point, we examine the sign of $f'(x)$ on either side of $x = -\frac{1}{5}$.

• For $x < -\frac{1}{5}$: Choose a test point like $x = -1$. $f'(-1) = e^{-5}(1 + 5(-1)) = e^{-5}(-4) < 0$ The derivative is negative, meaning the function is decreasing.

• For $x > -\frac{1}{5}$: Choose a test point like $x = 0$. $f'(0) = e^{0}(1 + 5(0)) = 1 > 0$ The derivative is positive, meaning the function is increasing.

Since the function changes from decreasing to increasing at $x = -\frac{1}{5}$, this indicates a local minimum at $x = -\frac{1}{5}$.

Step 4: Find the value of the function at this critical point

To find the value of the function at $x = -\frac{1}{5}$, we substitute this value into the original function:

$f\left(-\frac{1}{5}\right) = -\frac{1}{5} e^{5(-\frac{1}{5})} - 3$

$f\left(-\frac{1}{5}\right) = -\frac{1}{5} e^{-1} - 3$

Using the approximation $e^{-1} \approx 0.3679$:

$f\left(-\frac{1}{5}\right) \approx -\frac{1}{5} \cdot 0.3679 - 3$

$f\left(-\frac{1}{5}\right) \approx -0.07358 - 3 = -3.07358$

So, the value of the function at the local minimum is approximately $-3.07358$.

Final Answer:

• The location of the local minimum is at $x = -\frac{1}{5}$.
• The value of the function at this minimum is approximately $f\left(-\frac{1}{5}\right) \approx -3.07358$.

Would you like further clarification on any part of the solution? Or do you have any other questions on this topic?

Follow-up Questions:

1. How do you apply the second derivative test to verify the nature of critical points?
2. What are some other techniques for finding extrema of functions?
3. How can we analyze the behavior of functions at infinity?
4. How do you interpret inflection points and their role in graphing functions?
5. What are some real-world applications of using the first derivative test?

Tip: For the first derivative test, always check the sign of the derivative on either side of the critical point to determine whether it is a local minimum or maximum.