m(x)=x/8-3ln(4x+2) find maximum and minimum value

To find the maximum and minimum values of the function \( m(x) = \frac{x}{8} - 3 \ln(4x + 2) \), we will follow these steps:

### Step 1: Find the derivative of \( m(x) \)

To locate critical points (potential maxima or minima), we first find the derivative of \( m(x) \) and set it equal to zero.

1. Derivative of \( \frac{x}{8} \) is \( \frac{1}{8} \).
2. Derivative of \( -3 \ln(4x + 2) \):
   Using the chain rule:
   \[
   \frac{d}{dx} \left( -3 \ln(4x + 2) \right) = -3 \cdot \frac{1}{4x + 2} \cdot 4 = -\frac{12}{4x + 2}
   \]

Thus, the derivative of \( m(x) \), \( m'(x) \), is:
\[
m'(x) = \frac{1}{8} - \frac{12}{4x + 2}
\]

### Step 2: Set the derivative equal to zero

Set \( m'(x) = 0 \) to find the critical points:
\[
\frac{1}{8} - \frac{12}{4x + 2} = 0
\]

Multiply through by \( 8(4x + 2) \) to eliminate the denominators:
\[
(4x + 2) - 96 = 0
\]
\[
4x + 2 = 96
\]
\[
4x = 94
\]
\[
x = \frac{94}{4} = 23.5
\]

### Step 3: Check the second derivative or analyze intervals

To determine whether \( x = 23.5 \) is a maximum or minimum, we analyze the second derivative or use a sign chart. We can also plug in values around \( x = 23.5 \) into \( m'(x) \) to check whether it changes from positive to negative (indicating a maximum) or negative to positive (indicating a minimum).

### Step 4: Find the corresponding value of \( m(x) \)

Now, substitute \( x = 23.5 \) back into \( m(x) \):
\[
m(23.5) = \frac{23.5}{8} - 3 \ln(4(23.5) + 2)
\]
Simplifying:
\[
m(23.5) = \frac{23.5}{8} - 3 \ln(96)
\]
\[
m(23.5) = 2.9375 - 3 \ln(96)
\]
Now, approximate \( \ln(96) \):
\[
\ln(96) \approx 4.564
\]
Thus:
\[
m(23.5) \approx 2.9375 - 3 \times 4.564
\]
\[
m(23.5) \approx 2.9375 - 13.692
\]
\[
m(23.5) \approx -10.7545
\]

So, \( x = 23.5 \) yields a minimum value of approximately \( m(23.5) \approx -10.7545 \).

### Conclusion:

- The function \( m(x) \) has a **minimum value** at \( x = 23.5 \), with \( m(23.5) \approx -10.7545 \).
- There is no global maximum as \( m(x) \) increases without bound as \( x \to \infty \).

---

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

Here are five related questions to consider:
1. How do we analyze the behavior of a logarithmic function for large values of \( x \)?
2. Why does the derivative help us find maxima and minima?
3. What is the significance of the second derivative in determining concavity?
4. Can we determine if this minimum is global or local using the first derivative test?
5. What are the domain restrictions for \( m(x) \) based on the logarithmic term?

**Tip:** When differentiating logarithmic functions, always remember to apply the chain rule correctly, as logarithms often involve a nested function (like \( 4x + 2 \) here).

m(x) = x/8 - 3ln(4x + 2) find maximum and minimum value

Explore how to find the critical points, maximum, and minimum values for the function m(x) = x/8 - 3ln(4x + 2) using calculus techniques such as the chain rule and first/second derivative tests. This problem covers logarithmic differentiation and optimization, suitable for students in Grades 11-12.

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