Find A And B Such That F Is Differentiable Everywhere

The pursuit of differentiability in functions is a cornerstone of calculus, demanding a delicate balance of smoothness and continuity. When tasked with finding specific parameters that ensure a function is differentiable everywhere, we embark on a journey that blends algebraic manipulation with deep conceptual understanding. This exploration delves into the methods, considerations, and nuances involved in determining the values of constants, typically denoted as 'a' and 'b,' that guarantee a function's differentiability across its entire domain.

The Essence of Differentiability

Before diving into the methods of finding 'a' and 'b,' it's crucial to understand what differentiability truly means. A function f(x) is said to be differentiable at a point x = c if its derivative, f'(x), exists at that point. This implies two critical conditions:

• Continuity: The function must be continuous at x = c. This means that there is no break or jump in the graph of the function at that point. Mathematically, this is expressed as:

  lim (x→c-) f(x) = lim (x→c+) f(x) = f(c)
  

  where lim (x→c-) represents the limit as x approaches c from the left, and lim (x→c+) represents the limit as x approaches c from the right.

• Smoothness: The function must be "smooth" at x = c. This means that the left-hand derivative must equal the right-hand derivative at that point. In other words, the slope of the tangent line approaching x = c from the left must be the same as the slope of the tangent line approaching x = c from the right. Mathematically, this is expressed as:

  lim (h→0-) [f(c + h) - f(c)] / h = lim (h→0+) [f(c + h) - f(c)] / h
  

  or equivalently,

  f'(c-) = f'(c+)
  

  where f'(c-) represents the left-hand derivative at x = c, and f'(c+) represents the right-hand derivative at x = c.

General Steps to Determine 'a' and 'b'

When a function is defined piecewise, the points where the pieces connect are the most critical for examining differentiability. Here’s a step-by-step approach to finding 'a' and 'b' to ensure differentiability everywhere:

1. Identify the Piecewise Function: Express the function f(x) in its piecewise form. This involves identifying the intervals over which different expressions define the function.

2. Ensure Continuity: At each point where the function definition changes (the "connecting points"), ensure that the function is continuous. This is done by setting the values of the adjacent pieces equal to each other at that point. This step will typically give you one equation involving 'a' and 'b'.

3. Ensure Smoothness (Differentiability): Take the derivative of each piece of the function. Then, at each connecting point, set the derivatives of the adjacent pieces equal to each other. This ensures that the slopes of the tangent lines match at these points, guaranteeing a smooth transition. This step will typically give you another equation involving 'a' and 'b'.

4. Solve the System of Equations: Solve the system of equations generated in steps 2 and 3 to find the values of 'a' and 'b'. These values will ensure that the function is both continuous and differentiable at the connecting points, and thus differentiable everywhere.

Example: A Detailed Walkthrough

Let's consider a concrete example to illustrate the process. Suppose we have the following piecewise function:

f(x) =
  ax^2 + b,   if x ≤ 1
  x^3,       if x > 1

Our goal is to find the values of 'a' and 'b' such that f(x) is differentiable everywhere.

Step 1: Identify the Piecewise Function

The function is already given in piecewise form, with the two pieces defined as ax^2 + b for x ≤ 1 and x^3 for x > 1. The connecting point is x = 1.

Step 2: Ensure Continuity

For f(x) to be continuous at x = 1, the values of the two pieces must be equal at x = 1. Thus:

a(1)^2 + b = (1)^3
a + b = 1  --- (Equation 1)

Step 3: Ensure Smoothness (Differentiability)

First, find the derivative of each piece:

• For x ≤ 1: f'(x) = 2ax
• For x > 1: f'(x) = 3x^2

Now, for f(x) to be differentiable at x = 1, the derivatives of the two pieces must be equal at x = 1. Thus:

2a(1) = 3(1)^2
2a = 3
a = 3/2  --- (Equation 2)

Step 4: Solve the System of Equations

We now have a system of two equations:

a + b = 1
a = 3/2

Substituting the value of 'a' from Equation 2 into Equation 1:

3/2 + b = 1
b = 1 - 3/2
b = -1/2

Therefore, a = 3/2 and b = -1/2. With these values, the function f(x) is differentiable everywhere.

Verification

Let's verify our solution. With a = 3/2 and b = -1/2, the function becomes:

f(x) =
  (3/2)x^2 - 1/2,   if x ≤ 1
  x^3,             if x > 1

At x = 1:

• (3/2)(1)^2 - 1/2 = 3/2 - 1/2 = 1
• (1)^3 = 1

The function is continuous at x = 1.

The derivatives are:

• For x ≤ 1: f'(x) = 3x
• For x > 1: f'(x) = 3x^2

At x = 1:

• 3(1) = 3
• 3(1)^2 = 3

The derivatives are equal at x = 1. Therefore, the function is differentiable everywhere.

Common Challenges and Considerations

While the process seems straightforward, there are nuances that require careful attention:

• More Complex Functions: When dealing with trigonometric functions, exponential functions, or other more complex expressions, the derivatives can become more challenging to compute. Ensure accuracy in differentiation.
• Multiple Connecting Points: If the piecewise function has multiple connecting points, you will need to ensure continuity and differentiability at each of them, leading to a larger system of equations.
• Absolute Value Functions: Absolute value functions introduce sharp corners, making them non-differentiable at the point where the expression inside the absolute value equals zero. These require special handling, often involving rewriting the absolute value function as a piecewise function.
• Higher-Order Differentiability: Sometimes, a problem may require not just differentiability, but also that the second derivative (or higher) exists and is continuous. This would require equating higher-order derivatives at the connecting points.
• Implicit Differentiation: For functions defined implicitly, implicit differentiation must be used to find the derivatives needed to ensure smoothness at the connection points.
• Understanding Limits: A thorough understanding of limits is crucial for confirming continuity and differentiability, especially at points where the function's definition changes.

Examples of Complex Scenarios

1. Absolute Value Function:

Consider the function:

f(x) =
  x^2,          if x ≤ 1
  a|x - 2| + b,  if x > 1

To ensure differentiability, we first rewrite the absolute value function:

|x - 2| =
  -(x - 2),  if x < 2
  (x - 2),   if x ≥ 2

So, the function becomes:

f(x) =
  x^2,          if x ≤ 1
  -a(x - 2) + b, if 1 < x < 2
  a(x - 2) + b,  if x ≥ 2

We have two connection points: x = 1 and x = 2.

• At x = 1 (Continuity):

  (1)^2 = -a(1 - 2) + b
  1 = a + b  --- (Equation 1)
  

• At x = 1 (Differentiability): The derivatives are:

  • f'(x) = 2x for x ≤ 1
  • f'(x) = -a for 1 < x < 2

  So, at x = 1:

  2(1) = -a
  a = -2  --- (Equation 2)
  

• Solving for b: Substituting a = -2 into Equation 1:

  1 = -2 + b
  b = 3
  

• Checking x = 2: The function pieces are:

  • f(x) = -2(x - 2) + 3, if 1 < x < 2
  • f(x) = -2(x - 2) + 3, if x ≥ 2

  Which simplifies to:

  • f(x) = -2x + 7, if 1 < x < 2
  • f(x) = -2x + 7, if x ≥ 2

  There is no change in the function's definition at x=2, so the function will be differentiable at this point.

Therefore, a = -2 and b = 3 ensures differentiability everywhere.

2. Trigonometric Function:

Consider the function:

f(x) =
  a sin(x),  if x ≤ π/2
  b - cos(x), if x > π/2

• At x = π/2 (Continuity):

  a sin(π/2) = b - cos(π/2)
  a(1) = b - 0
  a = b  --- (Equation 1)
  

• At x = π/2 (Differentiability): The derivatives are:

  • f'(x) = a cos(x) for x ≤ π/2
  • f'(x) = sin(x) for x > π/2

  So, at x = π/2:

  a cos(π/2) = sin(π/2)
  a(0) = 1
  0 = 1
  

This leads to a contradiction, indicating that there are no values of 'a' and 'b' that make this function differentiable at x = π/2. This arises because the derivative of sin(x) at π/2 is 1, while the derivative of a sin(x) at π/2 can only be zero, regardless of the value of a.

3. Cubic and Quadratic Polynomials

Consider the function:

f(x) =
    x^3,            if x <= 0
    ax^2 + bx,      if x > 0

• At x = 0 (Continuity):

  (0)^3 = a(0)^2 + b(0)
  0 = 0
  

  This condition is already satisfied, and provides no constraint on a or b.

• At x = 0 (Differentiability):

  The derivatives are:

  • f'(x) = 3x^2 for x <= 0
  • f'(x) = 2ax + b for x > 0

  So, at x = 0:

  3(0)^2 = 2a(0) + b
  0 = b
  

  This gives us b = 0. The value of a is unconstrained, and can be any real number. Therefore, b = 0, and a can be any real number.

Conclusion

Finding values for constants that ensure a function is differentiable everywhere requires a solid grasp of continuity, differentiability, and the ability to solve systems of equations. While the basic process is relatively straightforward, the complexity increases with the intricacy of the functions involved. By systematically ensuring continuity and smoothness at each connecting point and carefully considering the types of functions, one can confidently determine the necessary values of 'a' and 'b' to achieve differentiability across the function's entire domain. Understanding the underlying principles and potential pitfalls allows for a robust approach to these problems, solidifying the foundation for more advanced calculus concepts. The exploration doesn't merely end with finding the values; it enriches the understanding of how functions behave and the critical conditions required for their smooth and predictable behavior.