Consider the function g defined by g (x) = e^2x - 2x, x ER.
Point A is the point on the graph of g for which the x-coordinate is In √3.
Find the equation of the tangent to the graph of g at point A.

Let's solve the problem step by step.

### Step 1: Define the Function
The given function is:

\[
g(x) = e^{2x} - 2x
\]

### Step 2: Find the Derivative of \( g(x) \)
To find the equation of the tangent line, we need the derivative of \( g(x) \) with respect to \( x \). The derivative, \( g'(x) \), gives us the slope of the tangent line at any point \( x \).

\[
g'(x) = \frac{d}{dx} \left( e^{2x} - 2x \right)
\]

The derivative of \( e^{2x} \) with respect to \( x \) is \( 2e^{2x} \), and the derivative of \( -2x \) with respect to \( x \) is \( -2 \).

So,

\[
g'(x) = 2e^{2x} - 2
\]

### Step 3: Find the Slope at Point A
Point A has an \( x \)-coordinate of \( x = \ln(\sqrt{3}) \).

To find the slope of the tangent at point A, substitute \( x = \ln(\sqrt{3}) \) into \( g'(x) \):

\[
g'\left(\ln(\sqrt{3})\right) = 2e^{2\ln(\sqrt{3})} - 2
\]

We can simplify \( e^{2\ln(\sqrt{3})} \) using the property \( e^{\ln(a^b)} = a^b \):

\[
e^{2\ln(\sqrt{3})} = (\sqrt{3})^2 = 3
\]

So the slope \( m \) at point A is:

\[
m = 2 \times 3 - 2 = 6 - 2 = 4
\]

### Step 4: Find the y-Coordinate of Point A
The \( y \)-coordinate of point A is found by substituting \( x = \ln(\sqrt{3}) \) into the original function \( g(x) \):

\[
g\left(\ln(\sqrt{3})\right) = e^{2\ln(\sqrt{3})} - 2\ln(\sqrt{3})
\]

We already found that \( e^{2\ln(\sqrt{3})} = 3 \), so:

\[
g\left(\ln(\sqrt{3})\right) = 3 - 2\ln(\sqrt{3})
\]

Thus, point A is \( \left(\ln(\sqrt{3}), 3 - 2\ln(\sqrt{3})\right) \).

### Step 5: Write the Equation of the Tangent Line
The equation of the tangent line at point A can be written as:

\[
y - y_1 = m(x - x_1)
\]

Substituting the coordinates of point A and the slope:

\[
y - \left(3 - 2\ln(\sqrt{3})\right) = 4\left(x - \ln(\sqrt{3})\right)
\]

### Step 6: Simplify the Equation
Expand and simplify the equation:

\[
y = 4x - 4\ln(\sqrt{3}) + 3 - 2\ln(\sqrt{3})
\]

Combine like terms:

\[
y = 4x - 6\ln(\sqrt{3}) + 3
\]

### Final Answer:
The equation of the tangent to the graph of \( g(x) = e^{2x} - 2x \) at point \( A \) is:

\[
y = 4x - 6\ln(\sqrt{3}) + 3
\]

Would you like more details on any part of the solution, or do you have any further questions?

Here are some related questions to consider:
1. How do you differentiate functions involving \( e^x \)?
2. What is the significance of the point-slope form of a line?
3. How do you simplify expressions involving logarithms?
4. How does the slope of the tangent line relate to the derivative of the function?
5. What are some common properties of exponential functions?

**Tip:** Always double-check your differentiation, especially with exponential functions, as small errors in the exponent can lead to incorrect results.

Learn how to find the equation of the tangent line to the graph of g(x) = e^(2x) - 2x at point A. This problem involves exponential functions, derivatives, and tangent lines. Suitable for advanced high school students or calculus learners.

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