If the function `f(x)=ax e^(-bx)` has a local maximum at the point (2,10), then

Options

A. a = 5e
B. a = 5
C. b = 1
D. b = 1/2

Explanation

To solve the problem, we need to find the values of \( a \) and \( b \) given that the function \( f(x) = ax e^{-bx} \) has a local maximum at the point \( (2, 10) \).

### Step-by-Step Solution:

**Step 1: Use the point (2, 10) to find an equation involving \( a \) and \( b \).**

Since the function has a local maximum at \( x = 2 \), we can substitute \( x = 2 \) into the function:

\[ f(2) = a(2)e^{-b(2)} = 10 \]

This simplifies to:

\[ 2a e^{-2b} = 10 \]

Dividing both sides by 2 gives:

\[ a e^{-2b} = 5 \quad \text{(Equation 1)} \]

**Step 2: Find the derivative of the function and set it to zero.**

To find the critical points, we need to take the derivative of \( f(x) \):

\[ f'(x) = \frac{d}{dx}(ax e^{-bx}) = a e^{-bx} + ax \frac{d}{dx}(e^{-bx}) \]

Using the product rule and the chain rule, we get:

\[ f'(x) = a e^{-bx} - abx e^{-bx} \]

Factoring out \( e^{-bx} \):

\[ f'(x) = e^{-bx}(a - abx) \]

Setting the derivative equal to zero for critical points:

\[ e^{-bx}(a - abx) = 0 \]

Since \( e^{-bx} \) is never zero, we have:

\[ a - abx = 0 \]

This implies:

\[ a(1 - bx) = 0 \]

Since \( a \neq 0 \) (as it would make the function zero), we have:

\[ 1 - bx = 0 \quad \Rightarrow \quad bx = 1 \quad \Rightarrow \quad b = \frac{1}{x} \]

At \( x = 2 \):

\[ b = \frac{1}{2} \quad \text{(Equation 2)} \]

**Step 3: Substitute \( b \) back into Equation 1 to find \( a \).**

Now we substitute \( b = \frac{1}{2} \) into Equation 1:

\[ a e^{-2(\frac{1}{2})} = 5 \]

This simplifies to:

\[ a e^{-1} = 5 \]

Thus,

\[ a = 5e \]

**Final Values:**

The values of \( a \) and \( b \) are:

\[ a = 5e \quad \text{and} \quad b = \frac{1}{2} \]

### Summary of the Solution:

- The function \( f(x) = ax e^{-bx} \) has a local maximum at \( (2, 10) \).
- From the conditions, we derived two equations.
- We found \( b = \frac{1}{2} \) and substituted back to find \( a = 5e \).