find the x value of points where hthe function defined as followed have any relative extrema and find the value of any relative extrema for f(x) = x^2 + 8x + 5

To find the relative extrema of the function \( f(x) = x^2 + 8x + 5 \), we need to take the following steps:

### Step 1: Differentiate the Function
First, find the first derivative of \( f(x) \) to determine the critical points. 

\[
f(x) = x^2 + 8x + 5
\]
Differentiating with respect to \( x \):
\[
f'(x) = 2x + 8
\]

### Step 2: Set the First Derivative Equal to Zero
To find the critical points, set \( f'(x) = 0 \) and solve for \( x \):

\[
2x + 8 = 0
\]
\[
2x = -8
\]
\[
x = -4
\]

So, the critical point is at \( x = -4 \).

### Step 3: Determine the Nature of the Critical Point
To confirm whether \( x = -4 \) is a minimum or maximum, we can use the second derivative test.

1. Calculate the second derivative:
   \[
   f''(x) = 2
   \]

2. Since \( f''(x) = 2 \) (which is positive), the function is concave up at \( x = -4 \), indicating a relative minimum.

### Step 4: Calculate the Value of the Relative Minimum
Now, substitute \( x = -4 \) back into the original function to find the minimum value:

\[
f(-4) = (-4)^2 + 8(-4) + 5
\]
\[
= 16 - 32 + 5
\]
\[
= -11
\]

### Conclusion
The function \( f(x) = x^2 + 8x + 5 \) has a relative minimum at \( x = -4 \) with a minimum value of \( f(-4) = -11 \).

---

Would you like further details on any part of this solution? 

Here are some related questions that build on this topic:

1. How do we determine concavity and its impact on relative extrema?
2. What is the second derivative test, and how does it work?
3. Can we identify relative maxima for functions similar to \( f(x) = x^2 + 8x + 5 \)?
4. How can we find inflection points in a function?
5. What is the difference between absolute and relative extrema?

**Tip:** When finding relative extrema, the first derivative identifies critical points, while the second derivative test reveals the nature of these points.

Find the x value of points where the function defined as f(x) = x^2 + 8x + 5 has any relative extrema and find the value of any relative extrema.

Discover how to find the x values where f(x) = x^2 + 8x + 5 has relative extrema. Learn to apply the first and second derivative tests, identify critical points, and determine the minimum value for this quadratic function. Ideal for high school calculus students.

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