The function f(x)=6x+9x−1f(x)=6x+9x^-1 has one local minimum and one local maximum.
This function has a local maximum at x=x=
with value

and a local minimum at x=x=
with value

To find the local minimum and maximum of the function \( f(x) = 6x + 9x^{-1} \), we need to determine the critical points by following these steps:

### Step 1: Differentiate \( f(x) \)
The function is:
\[
f(x) = 6x + \frac{9}{x}
\]

Taking the derivative \( f'(x) \):
\[
f'(x) = 6 - \frac{9}{x^2}
\]

### Step 2: Set \( f'(x) = 0 \) to find critical points
\[
6 - \frac{9}{x^2} = 0
\]
Solving for \( x \):
\[
\frac{9}{x^2} = 6
\]
\[
x^2 = \frac{9}{6} = \frac{3}{2}
\]
\[
x = \pm \sqrt{\frac{3}{2}}
\]

Thus, the critical points are \( x = \sqrt{\frac{3}{2}} \) and \( x = -\sqrt{\frac{3}{2}} \).

### Step 3: Determine if the critical points are maxima or minima
To classify these points, we can examine the second derivative \( f''(x) \) or use the first derivative test.

The second derivative of \( f(x) \) is:
\[
f''(x) = \frac{18}{x^3}
\]

Evaluating \( f''(x) \) at \( x = \sqrt{\frac{3}{2}} \) and \( x = -\sqrt{\frac{3}{2}} \):

1. For \( x = \sqrt{\frac{3}{2}} \):
   \[
   f''\left(\sqrt{\frac{3}{2}}\right) > 0
   \]
   This indicates a **local minimum**.

2. For \( x = -\sqrt{\frac{3}{2}} \):
   \[
   f''\left(-\sqrt{\frac{3}{2}}\right) < 0
   \]
   This indicates a **local maximum**.

### Step 4: Calculate \( f(x) \) at the critical points
1. **Local Maximum** at \( x = -\sqrt{\frac{3}{2}} \):
   \[
   f\left(-\sqrt{\frac{3}{2}}\right) = 6\left(-\sqrt{\frac{3}{2}}\right) + \frac{9}{-\sqrt{\frac{3}{2}}} = -12\sqrt{\frac{3}{2}}
   \]

2. **Local Minimum** at \( x = \sqrt{\frac{3}{2}} \):
   \[
   f\left(\sqrt{\frac{3}{2}}\right) = 6\sqrt{\frac{3}{2}} + \frac{9}{\sqrt{\frac{3}{2}}} = 12\sqrt{\frac{3}{2}}
   \]

### Conclusion
- **Local Maximum**: \( x = -\sqrt{\frac{3}{2}} \) with value \( -12\sqrt{\frac{3}{2}} \).
- **Local Minimum**: \( x = \sqrt{\frac{3}{2}} \) with value \( 12\sqrt{\frac{3}{2}} \).

Would you like further details or have any questions?

Here are five related questions for practice:
1. How do you determine the concavity of a function at a critical point?
2. What does the second derivative test indicate about a function's behavior?
3. How would the values of local maxima and minima change if the function had a different coefficient?
4. Can there be more than one local maximum or minimum in a polynomial of degree 3?
5. How can we interpret the behavior of \( f(x) \) as \( x \to 0 \) or \( x \to \infty \)?

**Tip:** Checking the sign of \( f''(x) \) at critical points quickly identifies local minima and maxima by indicating concavity.

The function f(x) = 6x + 9x^-1 has one local minimum and one local maximum. This function has a local maximum at x = with value and a local minimum at x = with value.

Learn how to determine the local maximum and minimum of the function f(x) = 6x + 9x^-1 through calculus techniques. This detailed solution explains how to find and classify critical points using the first and second derivative tests, ideal for Grade 12 or early college students.

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