Prove that the curve `y=e^(|x|)` cannot have a unique tangent line at the point `x = 0.` Find the angle between the one-sided tangents to the curve at the point `x = 0.`

Options

A. `(pi)/(4) `
B. `(pi)/(6) `
C. `(pi)/(2) `
D. `(pi)/(3) `

Explanation

To prove that the curve \( y = e^{|x|} \) cannot have a unique tangent line at the point \( x = 0 \) and to find the angle between the one-sided tangents to the curve at that point, we can follow these steps:

### Step 1: Define the function
The curve is defined as:
\[ y = e^{|x|} \]
This can be expressed piecewise as:
\[ y = \begin{cases} e^x & \text{if } x \geq 0 \\ e^{-x} & \text{if } x < 0 \end{cases} \]

### Step 2: Differentiate the function
To find the slope of the tangent lines, we need to differentiate \( y \) with respect to \( x \).

For \( x \geq 0 \):
\[ \frac{dy}{dx} = e^x \]

For \( x < 0 \):
\[ \frac{dy}{dx} = -e^{-x} \]

### Step 3: Evaluate the derivative at \( x = 0 \)
Now we will evaluate the derivative from both sides at \( x = 0 \).

- Right-hand derivative (as \( x \) approaches 0 from the right):
\[ \frac{dy}{dx} \bigg|_{x=0^+} = e^0 = 1 \]

- Left-hand derivative (as \( x \) approaches 0 from the left):
\[ \frac{dy}{dx} \bigg|_{x=0^-} = -e^0 = -1 \]

### Step 4: Conclusion about the uniqueness of the tangent line
Since the right-hand derivative at \( x = 0 \) is \( 1 \) and the left-hand derivative is \( -1 \), the slopes of the tangents from either side are different. Therefore, the curve cannot have a unique tangent line at the point \( x = 0 \).

### Step 5: Find the equations of the tangent lines
Using the point-slope form of the equation of a line \( y - y_1 = m(x - x_1) \):

- For the right-hand tangent (slope = 1):
\[ y - 1 = 1(x - 0) \implies y = x + 1 \]

- For the left-hand tangent (slope = -1):
\[ y - 1 = -1(x - 0) \implies y = -x + 1 \]

### Step 6: Find the angle between the two tangents
The slopes of the tangents are \( m_1 = 1 \) and \( m_2 = -1 \). The angle \( \theta \) between the two lines can be found using the formula:
\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting the values:
\[ \tan(\theta) = \left| \frac{1 - (-1)}{1 + (1)(-1)} \right| = \left| \frac{2}{0} \right| \]
Since the denominator is zero, this indicates that the lines are perpendicular, and thus:
\[ \theta = 90^\circ \quad \text{or} \quad \theta = \frac{\pi}{2} \text{ radians} \]

### Final Answer
The angle between the one-sided tangents to the curve at the point \( x = 0 \) is \( \frac{\pi}{2} \) radians.

---