Find the critical point(s) of the function 𝑓(𝑥)=(12−6𝑥)13𝑥f(x)=(12−6x)13x, if any, leaving your answer(s) in exact form (no approximations). Enter DNE if there are no critical points.

Answer: Critical point(s) =  

To find the critical points of the function \( f(x) = \frac{12 - 6x}{13x} \), we follow these steps:

### Step 1: Differentiate the function

To find critical points, we first need to take the derivative of the function. Using the quotient rule for derivatives, where if \( g(x) = \frac{u(x)}{v(x)} \), the derivative is:

\[
g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}
\]

In this case:
- \( u(x) = 12 - 6x \) and \( v(x) = 13x \).
- \( u'(x) = -6 \) and \( v'(x) = 13 \).

Now, applying the quotient rule:

\[
f'(x) = \frac{(-6)(13x) - (12 - 6x)(13)}{(13x)^2}
\]

### Step 2: Simplify the derivative

First, expand the numerator:

\[
f'(x) = \frac{-78x - 13(12 - 6x)}{(13x)^2}
\]

Distribute \( 13 \) over \( (12 - 6x) \):

\[
f'(x) = \frac{-78x - (156 - 78x)}{(13x)^2}
\]

Simplify the numerator:

\[
f'(x) = \frac{-78x - 156 + 78x}{(13x)^2}
\]

The \( -78x + 78x \) terms cancel out, so we are left with:

\[
f'(x) = \frac{-156}{(13x)^2}
\]

### Step 3: Find critical points

Critical points occur where \( f'(x) = 0 \) or where \( f'(x) \) is undefined. 

1. **Setting \( f'(x) = 0 \):**

Since the numerator of \( f'(x) \) is \( -156 \), and this is a constant, the derivative can never equal zero. Hence, there are no critical points where \( f'(x) = 0 \).

2. **Finding where \( f'(x) \) is undefined:**

The derivative \( f'(x) = \frac{-156}{(13x)^2} \) is undefined when the denominator is zero. This happens when \( x = 0 \).

However, \( x = 0 \) is not in the domain of the original function \( f(x) = \frac{12 - 6x}{13x} \), as division by zero is undefined.

### Conclusion:

Since the derivative does not equal zero anywhere, and the function is undefined at \( x = 0 \), **there are no critical points**.

Answer: **DNE** (Does Not Exist)

Would you like more details or have any questions?

Here are 5 related questions to expand on this:
1. How do we apply the quotient rule in more complex functions?
2. Why do critical points occur when the derivative equals zero?
3. Can there be critical points where the derivative is undefined but the function is continuous?
4. What is the significance of the domain when analyzing critical points?
5. How do second derivatives relate to finding concavity at critical points?

**Tip**: Always check both where the derivative equals zero and where it is undefined when looking for critical points!

Find the critical point(s) of the function f(x) = (12 - 6x)/(13x), if any, leaving your answer(s) in exact form (no approximations). Enter DNE if there are no critical points.

Explore how to determine the critical points of the function f(x) = (12 - 6x)/(13x) using the quotient rule for derivatives. This problem walks through differentiation, simplification, and identification of points where the function's derivative is zero or undefined, ideal for Grade 12 calculus students.

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