Problem 76 - Calculus Single Variable

Question

Suppose that \(f(x)=\left\\{\begin{array}{cl}e^{a x} & \text { if } \quad x \leq 0 \\ 1+\sin (b x) & \text { if } \quad x>0\end{array} .\right.\) How must we choose \(a\) and \(b\) so that \(f\) is a differentiable function?

Step-by-Step Solution

Answer

Choose \(a = b\) to make the function differentiable at \(x = 0\).

Step 1: Defining Differentiability

A function is differentiable at a point if it is continuous there and if the left-hand derivative equals the right-hand derivative. Here, we need to ensure that the function is differentiable at the point where the definition of the function changes, which is at \(x = 0\).

Step 2: Checking Continuity at x=0

For continuity at \(x = 0\), both portions of the function \(e^{ax}\) and \(1 + \sin(bx)\) must equal when \(x = 0\).Plug \(x = 0\) into both expressions:1. \(e^{a \cdot 0} = e^0 = 1\)2. \(1 + \sin(b \cdot 0) = 1 + \sin(0) = 1\)Both are equal, so the function is continuous at \(x = 0\).

Step 3: Finding the Left-Hand Derivative at x=0

Calculate the derivative of \(f(x) = e^{ax}\) when \(x \leq 0\).The derivative \(f'(x) = a e^{ax}\).Plugging in \(x = 0\):\[-f'(0^-) = a e^{a \cdot 0} = a \cdot 1 = a\]

Step 4: Finding the Right-Hand Derivative at x=0

Find the derivative of \(f(x) = 1 + \sin(bx)\) for \(x > 0\) which is \(f'(x) = b\cos(bx)\).Plug in \(x = 0\):\[f'(0^+) = b\cos(b\cdot 0) = b\cdot 1 = b\]

Step 5: Ensuring Equal Derivatives

The left-hand and right-hand derivatives at \(x=0\) must be equal for the function to be differentiable. Thus, \(f'(0^-) = f'(0^+)\) leads to\[a = b\]

Step 6: Conclusion

To ensure differentiability of \(f(x)\) at \(x = 0\), choose \(a = b\). This guarantees both continuity and equal derivatives at the junction.

Key Concepts

Continuity

To understand the concept of continuity at a specific point, think of a smooth path that has no breaks, jumps, or gaps. For a function to be continuous at a point, the value of the function must be the same from all approaching directions. In our exercise, continuity is examined at the point where the function switches definitions, specifically at \(x = 0\).

Continuity means:

• The left-side limit as \(x\) approaches 0 from the negative direction \((-x)\) is equal to the right-side limit as \(x\) approaches 0 from the positive direction
• These limits must equal the actual value of the function at that point

For the given function \(f(x)=\left\{\begin{array}{cl}e^{a x} & \text { if } \quad x \leq 0 \ 1+\sin (b x) & \text { if } \quad x>0\end{array} .\right.\), we checked and concluded it is continuous at \(x = 0\) because both expressions for the function yield 1 when \(x = 0\).

Left-hand derivative

The left-hand derivative refers to finding the derivative of the function as we approach a particular point from the left side. For our function, we need to consider the portion defined for \(x \leq 0\), which is \(f(x) = e^{ax}\).

To find this derivative at the point \(x = 0\):

• Differentiate the expression \(e^{ax}\) with respect to \(x\), yielding \(f'(x) = a e^{ax}\)
• Substitute into this derivative the value \(x = 0\)
• The result is \(f'(0^-) = a e^{a \cdot 0} = a \cdot 1 = a\)

This process tells us how the function behaves as we approach 0 from the left and provides a rate of change or slope from that direction.

Right-hand derivative

Next, we focus on the right-hand derivative, which considers the change in the function from the right side of the specific point. For our function, we look at the portion for \(x > 0\), expressed as \(f(x) = 1 + \sin(bx)\).

To find this derivative at \(x = 0\):

• Compute the derivative of \(1 + \sin(bx)\), giving \(f'(x) = b\cos(bx)\)
• Substitute \(x = 0\) into this expression
• The answer becomes \(f'(0^+) = b\cos(b \cdot 0) = b \cdot 1 = b\)

This represents the rate of change or slope of the function as it approaches 0 from the right.

Equal derivatives

For a function to be differentiable at a point, one crucial condition is that the left-hand derivative must equal the right-hand derivative at that point. This means the slope of the function from the left must match the slope from the right, resulting in a smooth transition without any sharp turns or cusps.

In our exercise, this condition is captured by setting \(f'(0^-) = f'(0^+)\). This requirement leads us to the equation \(a = b\). When both derivatives are equal, which happens when \(a\) equals \(b\), it suggests that the function transitions seamlessly through \(x=0\), maintaining the same rate of change from both sides.

Meeting this condition ensures that the function is not only continuous but also differentiable at the point \(x = 0\). This is why, in conclusion, to achieve differentiability at this point, it's necessary for \(a = b\).