To find the equation of the tangent line for the given functions at the point $(0,1)$, let's work through each part step by step.

(a) $y = e^{2x}$

Step 1: Find the derivative

The derivative of $y = e^{2x}$ is: $y'(x) = \frac{d}{dx} \left( e^{2x} \right) = 2e^{2x}.$

Step 2: Evaluate the slope at $x = 0$

The slope of the tangent line at $x = 0$ is: $y'(0) = 2e^{2(0)} = 2.$

Step 3: Equation of the tangent line

The tangent line has the equation: $y - y_1 = m(x - x_1),$ where $m = 2$ and $(x_1, y_1) = (0, 1)$. Substituting these values: $y - 1 = 2(x - 0).$

Simplify: $y = 2x + 1.$

(b) $y = e^{-2x}$

Step 1: Find the derivative

The derivative of $y = e^{-2x}$ is: $y'(x) = \frac{d}{dx} \left( e^{-2x} \right) = -2e^{-2x}.$

Step 2: Evaluate the slope at $x = 0$

The slope of the tangent line at $x = 0$ is: $y'(0) = -2e^{-2(0)} = -2.$

Step 3: Equation of the tangent line

Using the same formula for the tangent line: $y - y_1 = m(x - x_1),$ where $m = -2$ and $(x_1, y_1) = (0, 1)$. Substituting these values: $y - 1 = -2(x - 0).$

Simplify: $y = -2x + 1.$

Final Results:

• (a) $y(x) = 2x + 1$
• (b) $y(x) = -2x + 1$

Do you want further clarification or additional steps?

Expanding Questions:

1. How do derivatives help in finding tangent lines?
2. What happens if we shift the point of tangency to a different point, like $(1, e^2)$?
3. Why is the slope different for $e^{2x}$ and $e^{-2x}$ despite their exponential nature?
4. Can the method be extended to higher-dimensional tangent planes for functions of multiple variables?
5. What is the geometrical interpretation of the negative slope in $y = e^{-2x}$?

Tip:

Always check your derivative and substitution steps carefully to avoid calculation errors!