To find the slope and the equation of the tangent line to the graph of $f(x) = e^x + \frac{3}{e^x}$ at $x = 0$, we need to follow these steps:

Step 1: Find $f'(x)$

The slope of the tangent line at a point $x$ on the graph is given by the derivative $f'(x)$ evaluated at that point.

The function is: $f(x) = e^x + \frac{3}{e^x}$

The derivative $f'(x)$ is calculated by differentiating each term:

1. The derivative of $e^x$ is $e^x$.
2. The derivative of $\frac{3}{e^x} = 3 \cdot e^{-x}$ is $-3e^{-x}$.

So, $f'(x) = e^x - \frac{3}{e^x}$

Step 2: Evaluate $f'(0)$ for the slope $m$

Now, substitute $x = 0$ into $f'(x)$ to find the slope of the tangent line: $f'(0) = e^0 - \frac{3}{e^0} = 1 - 3 = -2$

Thus, the slope of the tangent line is: $m = -2$

Step 3: Find the equation of the tangent line

To write the equation of the tangent line, we use the point-slope form: $y - f(0) = m(x - 0)$

First, calculate $f(0)$: $f(0) = e^0 + \frac{3}{e^0} = 1 + 3 = 4$

So, the point at $x = 0$ is $(0, 4)$, and we have the slope $m = -2$.

Using the point-slope form: $y - 4 = -2(x - 0)$ which simplifies to: $y = -2x + 4$

Final Answer

• Slope of the tangent line: $m = -2$
• Equation of the tangent line: $y = -2x + 4$

Would you like further details or have any questions?

Here are some related questions to deepen understanding:

1. How would the tangent line change if the function were $f(x) = e^x - \frac{3}{e^x}$ instead?
2. What is the significance of finding the tangent line at a specific point?
3. How does the exponential function affect the shape and slope of the graph?
4. How would you find the second derivative $f''(x)$ to analyze concavity?
5. What happens to the tangent line equation if we evaluate it at a different point, like $x = 1$?

Tip: For exponential functions, remember that derivatives can often involve both positive and negative exponents, reflecting growth and decay in different directions.