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The question in the image asks to work with the function \( g(x) = \frac{e^{3x} + x}{2} \).

### a) Determine \( g'(x) \) for any real \( x \).

To find the derivative \( g'(x) \), we'll apply the standard rules of differentiation.

1. Given:
   \[
   g(x) = \frac{e^{3x} + x}{2}
   \]

2. Differentiate both terms in the numerator:
   - The derivative of \( e^{3x} \) with respect to \( x \) is \( 3e^{3x} \) (using the chain rule).
   - The derivative of \( x \) is \( 1 \).

3. So, the derivative of the numerator \( e^{3x} + x \) is \( 3e^{3x} + 1 \).

4. Since the entire function is divided by 2, we divide the derivative by 2:
   \[
   g'(x) = \frac{3e^{3x} + 1}{2}
   \]

### b) Determine the equation of the tangent line to the curve at the point where \( x = 0 \).

To find the equation of the tangent line at \( x = 0 \), we need the following:

1. **Slope of the tangent line:** This is given by \( g'(0) \).
   - First, evaluate \( g'(x) \) at \( x = 0 \):
     \[
     g'(0) = \frac{3e^{3(0)} + 1}{2} = \frac{3e^0 + 1}{2} = \frac{3 + 1}{2} = 2
     \]
   So, the slope of the tangent line is \( 2 \).

2. **Point of tangency:** We need to find \( g(0) \).
   - Evaluate \( g(x) \) at \( x = 0 \):
     \[
    

g is the function defined on ℝ by g(x) = (e^(3x) + x)/2. a) Determine g'(x) for any real x. b) Determine the equation of the tangent to the curve at the point of abscissa 0.

This problem explores the differentiation of the function g(x) = (e^(3x) + x)/2. Learn how to calculate g'(x) and find the equation of the tangent line at x = 0. This exercise applies the chain rule and exponential functions, suitable for students in Grades 10-12.

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