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Let c > 0 be some constant and consider the plots of the functions f(x) = e

−cx and g(x) = e

cx below.

Let A = (a, f(a)) be the point where the tangent line of f goes through the origin and likewise B = (b, g(b)).

Find the equation of the tangent lines through A and B.

Hint: This requires solving for a, b but as a function of c.

To solve for the equations of the tangent lines that pass through the origin for the functions \( f(x) = e^{-cx} \) and \( g(x) = e^{cx} \), we need to:

1. Find the points \( A = (a, f(a)) \) and \( B = (b, g(b)) \), where the tangent lines of these functions pass through the origin. 
2. Determine the slopes of the tangent lines at \( x = a \) and \( x = b \), then solve for \( a \) and \( b \).

### Step 1: Tangent Line and Slope

For any function \( y = f(x) \), the equation of the tangent line at a point \( x = a \) can be written as:
\[
y - f(a) = f'(a)(x - a)
\]
This tangent line passes through the origin \( (0, 0) \), so we substitute \( x = 0 \) and \( y = 0 \) into the tangent line equation:
\[
0 - f(a) = f'(a)(0 - a)
\]
This simplifies to:
\[
-f(a) = -a f'(a)
\]
or equivalently:
\[
f(a) = a f'(a)
\]

### Step 2: Apply to \( f(x) = e^{-cx} \)

The derivative of \( f(x) = e^{-cx} \) is:
\[
f'(x) = -c e^{-cx}
\]

Now, use the condition \( f(a) = a f'(a) \):
\[
e^{-ca} = a (-c e^{-ca})
\]
Simplifying:
\[
e^{-ca} = -ac e^{-ca}
\]
Dividing both sides by \( e^{-ca} \) (which is non-zero):
\[
1 = -ac
\]
Thus:
\[
a = -\frac{1}{c}
\]

### Step 3: Apply to \( g(x) = e^{cx} \)

The derivative of \( g(x) = e^{cx} \) is:
\[
g'(x) = c e^{cx}
\]

Now, use the condition \( g(b) = b g'(b) \):
\[
e^{cb} = b (c e^{cb})
\]
Simplifying:
\[
e^{cb} = bc e^{cb}
\]
Dividing both sides by \( e^{cb} \) (which is non-zero):
\[
1 = bc
\]
Thus:
\[
b = \frac{1}{c}
\]

### Step 4: Equation of the Tangent Lines

Now that we have \( a = -\frac{1}{c} \) and \( b = \frac{1}{c} \), we can find the equations of the tangent lines at these points.

1. **Tangent line at \( A = (a, f(a)) \):**

The slope of the tangent line at \( A \) is \( f'(a) \):
\[
f'(a) = -c e^{-c(-1/c)} = -c e^1 = -ce
\]
Thus, the equation of the tangent line at \( A \) is:
\[
y - e^1 = -ce (x - (-\frac{1}{c}))
\]
Simplifying:
\[
y = -ce \left(x + \frac{1}{c}\right) + e
\]

2. **Tangent line at \( B = (b, g(b)) \):**

The slope of the tangent line at \( B \) is \( g'(b) \):
\[
g'(b) = c e^{c(1/c)} = c e^1 = ce
\]
Thus, the equation of the tangent line at \( B \) is:
\[
y - e^1 = ce (x - \frac{1}{c})
\]
Simplifying:
\[
y = ce \left(x - \frac{1}{c}\right) + e
\]

### Conclusion

- The equation of the tangent line through \( A = \left(-\frac{1}{c}, e\right) \) is:
\[
y = -ce \left(x + \frac{1}{c}\right) + e
\]
- The equation of the tangent line through \( B = \left(\frac{1}{c}, e\right) \) is:
\[
y = ce \left(x - \frac{1}{c}\right) + e
\]

Would you like further clarifications or details?

Here are 5 related questions:
1. How do the slopes of the tangent lines relate to the exponential functions \( f(x) \) and \( g(x) \)?
2. What is the geometric interpretation of the points \( A \) and \( B \) in terms of symmetry?
3. How would the solution change if the functions had different exponential bases?
4. Can you extend this method to find tangents for more general exponential functions, such as \( e^{kx} \)?
5. What happens to the tangent lines as \( c \) approaches 0 or infinity?

**Tip**: Understanding how to compute the slope of tangent lines using derivatives is essential for solving many problems involving curves and tangents in calculus.

Let c > 0 be some constant and consider the plots of the functions f(x) = e^{-cx} and g(x) = e^{cx}. Let A = (a, f(a)) be the point where the tangent line of f goes through the origin and likewise B = (b, g(b)). Find the equation of the tangent lines through A and B.

This page explains how to find the tangent lines of the functions f(x) = e^{-cx} and g(x) = e^{cx} at points A and B, where the tangent lines pass through the origin. It covers key calculus concepts, including derivative rules and tangent line equations. The solution includes step-by-step reasoning, making it suitable for undergraduate students learning calculus.

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