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To find the equation of the normal line to the function \( f(x) = x^{3/2} + x^e + e^x + 2 \) at \( x = 0 \), we will follow these steps:

1. **Evaluate \( f(0) \)** to get the point on the curve.
2. **Find the derivative \( f'(x) \)** to get the slope of the tangent line at \( x = 0 \).
3. Use the fact that the slope of the normal line is the negative reciprocal of the slope of the tangent line.
4. Use the point-slope form of the line equation to write the equation of the normal line.

### Step 1: Evaluate \( f(0) \)
\[
f(0) = 0^{3/2} + 0^e + e^0 + 2 = 0 + 0 + 1 + 2 = 3
\]
So, the point on the curve at \( x = 0 \) is \( (0, 3) \).

### Step 2: Find the derivative \( f'(x) \)
We differentiate each term of \( f(x) \):
- The derivative of \( x^{3/2} \) is \( \frac{3}{2}x^{1/2} \).
- The derivative of \( x^e \) is \( e x^{e-1} \) (using the power rule).
- The derivative of \( e^x \) is \( e^x \).
- The constant term 2 has a derivative of 0.

Thus, the derivative is:
\[
f'(x) = \frac{3}{2}x^{1/2} + e x^{e-1} + e^x
\]

### Step 3: Evaluate \( f'(0) \)
At \( x = 0 \):
- \( \frac{3}{2}x^{1/2} = \frac{3}{2} \cdot 0^{1/2} = 0 \),
- \( e x^{e-1} = e \cdot 0^{e-1} = 0 \),
- \( e^0 = 1 \).

Therefore, \( f'(0) = 1 \), meaning the slope of the tangent line is 1.

### Step 4: Find the slope of the normal line
The slope of the normal line is the negative reciprocal of 1, which is \( -1 \).

### Step 5: Write the equation of the normal line
Using the point-slope form \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point \( (0, 3) \):
\[
y - 3 = -1(x - 0)
\]
\[
y = -x + 3
\]

Thus, the equation of the normal line is:
\[
y = -x + 3
\]

Would you like more details or clarifications?

### Related Questions:
1. What is the equation of the tangent line to the function at \( x = 0 \)?
2. How do we interpret the geometric meaning of a normal line?
3. Can you find the second derivative \( f''(x) \) at \( x = 0 \)?
4. What happens to the normal line equation if we evaluate it at another point?
5. How do the normal lines behave for functions with higher powers, like \( x^5 \)?

### Tip:
The normal line is always perpendicular to the tangent line at a given point on the curve, and its slope is the negative reciprocal of the tangent's slope.

Find the equation of the normal line to f(x) = x^(3/2) + x^e + e^x + 2 at x = 0.

Learn how to find the equation of the normal line to the function f(x) = x^(3/2) + x^e + e^x + 2 at x = 0. This problem covers differentiation, the slope of the tangent line, and the normal line equation. Ideal for students studying calculus at Grades 11-12.

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