Math Problem Statement

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.

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.

Solution

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

  1. Find the points A=(a,f(a))A = (a, f(a)) and B=(b,g(b))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=ax = a and x=bx = b, then solve for aa and bb.

Step 1: Tangent Line and Slope

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

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

The derivative of f(x)=ecxf(x) = e^{-cx} is: f(x)=cecxf'(x) = -c e^{-cx}

Now, use the condition f(a)=af(a)f(a) = a f'(a): eca=a(ceca)e^{-ca} = a (-c e^{-ca}) Simplifying: eca=acecae^{-ca} = -ac e^{-ca} Dividing both sides by ecae^{-ca} (which is non-zero): 1=ac1 = -ac Thus: a=1ca = -\frac{1}{c}

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

The derivative of g(x)=ecxg(x) = e^{cx} is: g(x)=cecxg'(x) = c e^{cx}

Now, use the condition g(b)=bg(b)g(b) = b g'(b): ecb=b(cecb)e^{cb} = b (c e^{cb}) Simplifying: ecb=bcecbe^{cb} = bc e^{cb} Dividing both sides by ecbe^{cb} (which is non-zero): 1=bc1 = bc Thus: b=1cb = \frac{1}{c}

Step 4: Equation of the Tangent Lines

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

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

The slope of the tangent line at AA is f(a)f'(a): f(a)=cec(1/c)=ce1=cef'(a) = -c e^{-c(-1/c)} = -c e^1 = -ce Thus, the equation of the tangent line at AA is: ye1=ce(x(1c))y - e^1 = -ce (x - (-\frac{1}{c})) Simplifying: y=ce(x+1c)+ey = -ce \left(x + \frac{1}{c}\right) + e

  1. Tangent line at B=(b,g(b))B = (b, g(b)):

The slope of the tangent line at BB is g(b)g'(b): g(b)=cec(1/c)=ce1=ceg'(b) = c e^{c(1/c)} = c e^1 = ce Thus, the equation of the tangent line at BB is: ye1=ce(x1c)y - e^1 = ce (x - \frac{1}{c}) Simplifying: y=ce(x1c)+ey = ce \left(x - \frac{1}{c}\right) + e

Conclusion

  • The equation of the tangent line through A=(1c,e)A = \left(-\frac{1}{c}, e\right) is: y=ce(x+1c)+ey = -ce \left(x + \frac{1}{c}\right) + e
  • The equation of the tangent line through B=(1c,e)B = \left(\frac{1}{c}, e\right) is: y=ce(x1c)+ey = 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)f(x) and g(x)g(x)?
  2. What is the geometric interpretation of the points AA and BB 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 ekxe^{kx}?
  5. What happens to the tangent lines as cc 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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Exponential Functions
Derivatives
Tangents

Formulas

f(x) = e^{-cx}
g(x) = e^{cx}
f'(x) = -c e^{-cx}
g'(x) = c e^{cx}
Tangent Line Equation: y - f(a) = f'(a)(x - a)

Theorems

Derivative of Exponential Functions

Suitable Grade Level

Undergraduate Calculus