The function f(x)=$$ \left(6x-5\right)e^{-2x} $$has one critical number. find it. x=

To find the critical number of the function $ f(x) = (6x - 5)e^{-2x} $, we need to follow these steps:

1. **Find the derivative of the function** $ f(x) $.
2. **Set the derivative equal to zero** to find the critical point(s).
3. **Solve for** $ x $.

### Step 1: Find the derivative

Using the product rule to differentiate $ f(x) = (6x - 5)e^{-2x} $, we have:

\[
f'(x) = \frac{d}{dx} \left(6x - 5\right) \cdot e^{-2x} + (6x - 5) \cdot \frac{d}{dx} e^{-2x}
\]

We can now differentiate each part:

- The derivative of $ 6x - 5 $ is $ 6 $.
- The derivative of $ e^{-2x} $ is $ -2e^{-2x} $ (using the chain rule).

Now applying the product rule:

\[
f'(x) = 6e^{-2x} + (6x - 5)(-2e^{-2x})
\]

Simplify the expression:

\[
f'(x) = 6e^{-2x} - 2(6x - 5)e^{-2x}
\]

Factor out $ e^{-2x} $:

\[
f'(x) = e^{-2x} \left(6 - 2(6x - 5)\right)
\]

Simplify the expression inside the parentheses:

\[
f'(x) = e^{-2x} \left(6 - 12x + 10\right)
\]

\[
f'(x) = e^{-2x} \left(16 - 12x\right)
\]

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

To find the critical number, set $ f'(x) = 0 $:

\[
e^{-2x} (16 - 12x) = 0
\]

Since $ e^{-2x} $ is never zero, we only need to solve:

\[
16 - 12x = 0
\]

### Step 3: Solve for $ x $

\[
12x = 16
\]

\[
x = \frac{16}{12} = \frac{4}{3}
\]

Thus, the critical number is $ x = \frac{4}{3} $.

Let me know if you'd like further details or if you have any questions! Here are five related questions to expand this concept:

1. What are the second derivative tests for determining if a critical point is a maximum or minimum?
2. How does the chain rule work in differentiation, and how was it applied here?
3. How does the product rule work, and what are its typical applications in calculus?
4. What other techniques can be used to find critical points of functions?
5. Can this function have any inflection points? How would you find them?

**Tip:** Always check if a critical point is a maximum, minimum, or saddle point by using the second derivative test.

The function f(x) = (6x - 5)e^{-2x} has one critical number. Find it. x=

Learn how to find the critical number of the function f(x) = (6x - 5)e^{-2x}. This problem involves using the product rule and chain rule to differentiate and solve for the critical point. Suitable for advanced high school or college calculus students.

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