Minimum distance between the curves `f(x)=e^x and g(x)=ln x` is

Options

A. 1
B. `sqrt(2)`
C. 2
D. e

Answer

To find the minimum distance between the curves \( f(x) = e^x \) and \( g(x) = \ln x \), we can follow these steps:

### Step 1: Understand the Functions
The functions given are:
- \( f(x) = e^x \)
- \( g(x) = \ln x \)

These functions are inverses of each other. This means that if \( y = e^x \), then \( x = \ln y \).

### Step 2: Identify the Points of Interest
We need to find points on each curve where the distance between them is minimized. Let:
- Point \( P \) on curve \( f(x) \) be \( (x_1, f(x_1)) = (x_1, e^{x_1}) \)
- Point \( Q \) on curve \( g(x) \) be \( (x_2, g(x_2)) = (x_2, \ln x_2) \)

### Step 3: Find the Slopes
The slope of the tangent to the curve \( f(x) \) is given by its derivative:
\[ f'(x) = e^x \]
The slope of the tangent to the curve \( g(x) \) is:
\[ g'(x) = \frac{1}{x} \]

### Step 4: Set the Slopes Equal
Since the shortest distance occurs along the common normal, we set the slopes equal to the slope of the line \( y = x \), which has a slope of 1:
\[ e^{x_1} = 1 \quad \text{(for curve } f(x) \text{)} \]
\[ \frac{1}{x_2} = 1 \quad \text{(for curve } g(x) \text{)} \]

### Step 5: Solve for \( x_1 \) and \( x_2 \)
From \( e^{x_1} = 1 \):
\[ x_1 = 0 \quad \Rightarrow \quad f(0) = e^0 = 1 \quad \Rightarrow \quad P(0, 1) \]

From \( \frac{1}{x_2} = 1 \):
\[ x_2 = 1 \quad \Rightarrow \quad g(1) = \ln 1 = 0 \quad \Rightarrow \quad Q(1, 0) \]

### Step 6: Calculate the Distance
Now, we calculate the distance \( d \) between points \( P(0, 1) \) and \( Q(1, 0) \) using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(1 - 0)^2 + (0 - 1)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Conclusion
The minimum distance between the curves \( f(x) = e^x \) and \( g(x) = \ln x \) is \( \sqrt{2} \).

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