For the function `f(x)=1+3^x ln 3` find the antiderivative F(x), which assumes the value 7 for `x=2`. At what values of x does the curve `F(x)` cut the x-axis?

Options

A. x=3
B. x=1
C. x=0
D. None of these

Answer

To solve the problem, we will find the antiderivative \( F(x) \) of the function \( f(x) = 1 + 3^x \ln 3 \) and determine the values of \( x \) where the curve \( F(x) \) intersects the x-axis.

### Step 1: Find the antiderivative \( F(x) \)

We start with the function:
\[ f(x) = 1 + 3^x \ln 3 \]

To find the antiderivative \( F(x) \), we need to integrate \( f(x) \):
\[ F(x) = \int f(x) \, dx = \int (1 + 3^x \ln 3) \, dx \]

### Step 2: Integrate each term

1. The integral of \( 1 \) is:
\[ \int 1 \, dx = x \]

2. For the term \( 3^x \ln 3 \):
\[ \int 3^x \ln 3 \, dx = \frac{3^x}{\ln 3} \]

Putting it all together, we have:
\[ F(x) = x + \frac{3^x}{\ln 3} + C \]

### Step 3: Use the condition \( F(2) = 7 \) to find \( C \)

We know that \( F(2) = 7 \):
\[ F(2) = 2 + \frac{3^2}{\ln 3} + C = 7 \]
\[ 2 + \frac{9}{\ln 3} + C = 7 \]
\[ C = 7 - 2 - \frac{9}{\ln 3} \]
\[ C = 5 - \frac{9}{\ln 3} \]

### Step 4: Write the final form of \( F(x) \)

Substituting \( C \) back into the expression for \( F(x) \):
\[ F(x) = x + \frac{3^x}{\ln 3} + 5 - \frac{9}{\ln 3} \]

### Step 5: Find where \( F(x) \) cuts the x-axis

The curve \( F(x) \) cuts the x-axis when \( F(x) = 0 \):
\[ x + \frac{3^x}{\ln 3} + 5 - \frac{9}{\ln 3} = 0 \]

Rearranging gives:
\[ \frac{3^x}{\ln 3} = -x - 5 + \frac{9}{\ln 3} \]

This equation is transcendental and may require numerical methods or graphing to find the exact values of \( x \).

### Step 6: Approximate the value of \( x \)

To find the approximate values of \( x \) where \( F(x) = 0 \), we can test some values:

1. For \( x = 1 \):
\[ F(1) = 1 + \frac{3^1}{\ln 3} + 5 - \frac{9}{\ln 3} \]
This simplifies to:
\[ F(1) = 1 + \frac{3}{\ln 3} + 5 - \frac{9}{\ln 3} = 6 - \frac{6}{\ln 3} \]
Since \( \ln 3 \) is approximately \( 1.0986 \), we can see that \( F(1) \) is positive.

2. For \( x = 0 \):
\[ F(0) = 0 + \frac{3^0}{\ln 3} + 5 - \frac{9}{\ln 3} = \frac{1}{\ln 3} + 5 - \frac{9}{\ln 3} = 5 - \frac{8}{\ln 3} \]
This is also positive.

3. For \( x = -1 \):
\[ F(-1) = -1 + \frac{3^{-1}}{\ln 3} + 5 - \frac{9}{\ln 3} = -1 + \frac{1/3}{\ln 3} + 5 - \frac{9}{\ln 3} \]
This can be computed to check if it becomes negative.

Continuing this process will help find the approximate point where \( F(x) = 0 \).

### Summary

The antiderivative \( F(x) \) is:
\[ F(x) = x + \frac{3^x}{\ln 3} + 5 - \frac{9}{\ln 3} \]

To find where \( F(x) \) cuts the x-axis, we solve \( F(x) = 0 \).