Graph of `f(x)=x ` & `g(x)= 2+lnx` , x>0 where e is base of natural log. Graph intersect at=?

Options

A. once in `(0,1)` and never in `(1,oo)`
B. once in `(0,1)` and once in `(e ^(2) , oo)`
C. once in `(0,1)` and one in `(e, e ^(2))`
D. more than twice in `(0,oo)`

Answer

To find the points of intersection of the graphs of the functions \( f(x) = x \) and \( g(x) = 2 + \ln x \) for \( x > 0 \), we need to set the two functions equal to each other:

\[ f(x) = g(x) \]

This gives us the equation:

\[ x = 2 + \ln x \]

### Step 1: Rearranging the Equation
We can rearrange this equation to isolate the logarithmic term:

\[ x - 2 = \ln x \]

### Step 2: Analyzing the Functions
To analyze the intersection points, we can define a new function:

\[ h(x) = x - 2 - \ln x \]

We want to find the values of \( x \) for which \( h(x) = 0 \).

### Step 3: Finding the Derivative
Next, we can find the derivative of \( h(x) \) to understand its behavior:

\[ h'(x) = 1 - \frac{1}{x} \]

### Step 4: Critical Points
Setting the derivative to zero to find critical points:

\[ 1 - \frac{1}{x} = 0 \implies x = 1 \]

### Step 5: Analyzing the Sign of the Derivative
- For \( x < 1 \), \( h'(x) < 0 \) (decreasing).
- For \( x > 1 \), \( h'(x) > 0 \) (increasing).

This indicates that \( h(x) \) has a minimum at \( x = 1 \).

### Step 6: Evaluating \( h(1) \)
Now, we evaluate \( h(1) \):

\[ h(1) = 1 - 2 - \ln(1) = 1 - 2 - 0 = -1 \]

### Step 7: Evaluating \( h(2) \)
Next, we evaluate \( h(2) \):

\[ h(2) = 2 - 2 - \ln(2) = 0 - \ln(2) = -\ln(2) < 0 \]

### Step 8: Evaluating \( h(e) \)
Now, we evaluate \( h(e) \):

\[ h(e) = e - 2 - \ln(e) = e - 2 - 1 = e - 3 \]

Since \( e \approx 2.718 \), we find that \( h(e) > 0 \).

### Step 9: Conclusion on Intersections
Since \( h(1) < 0 \) and \( h(e) > 0 \), by the Intermediate Value Theorem, there is at least one root in the interval \( (1, e) \).

### Step 10: Finding the Second Intersection
To find the second intersection, we can evaluate \( h(x) \) at \( x = e^2 \):

\[ h(e^2) = e^2 - 2 - \ln(e^2) = e^2 - 2 - 2 = e^2 - 4 \]

Since \( e^2 \approx 7.389 \), we find that \( h(e^2) > 0 \). Therefore, there is a root in the interval \( (e, e^2) \).

### Final Result
Thus, the graphs of \( f(x) \) and \( g(x) \) intersect at two points: one in the interval \( (1, e) \) and another in the interval \( (e, e^2) \).