Understanding Differentiability for Piecewise Functions

For a function \(f(x)\) to be differentiable at a point, say \(x=c\), two main conditions must be met:

1. The function must be continuous at \(x=c\).
2. The left-hand derivative and the right-hand derivative at \(x=c\) must exist and be equal.

The given function is defined as:

\(f(x)=\left\{\begin{array}{cc} a x(x+1)+b, & x<1 \\ x-1, & 1 \leq x \leq 2 \end{array}\right.\)

We are told that the function \(f(x)\) is differentiable at \(x=1\). We need to find the value of \(a+b\).

Step 1: Ensure Continuity at the Point

For \(f(x)\) to be differentiable at \(x=1\), it must first be continuous at \(x=1\). This means the limit of the function as \(x\) approaches 1 from the left must equal the limit as \(x\) approaches 1 from the right, and both must be equal to the function's value at \(x=1\).

\(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)\)

Step 2: Calculate the Left-Hand and Right-Hand Limits

The left-hand limit uses the definition of \(f(x)\) for \(x < 1\):

\(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (a x(x+1)+b)\)

Substitute \(x=1\) into this expression:

\(a(1)(1+1)+b = a(1)(2)+b = 2a+b\)

The right-hand limit uses the definition of \(f(x)\) for \(x \geq 1\):

\(\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x-1)\)

Substitute \(x=1\) into this expression:

\(1-1 = 0\)

The function value at \(x=1\) also uses the definition for \(x \geq 1\):

\(f(1) = 1-1 = 0\)

Step 3: Set up the Continuity Equation

For continuity at \(x=1\), the limits and function value must be equal:

\(2a+b = 0 = 0\)

This gives us our first equation:

\(2a+b = 0 \quad (Equation\;1)\)

Step 4: Calculate the Derivatives for Each Piece

Now, we need to consider the differentiability condition. We find the derivative of each piece of the function.

For \(x < 1\), \(f(x) = ax(x+1)+b = ax^2+ax+b\). The derivative is:

\(f'(x) = \frac{d}{dx}(ax^2+ax+b) = 2ax+a\)

For \(1 < x < 2\), \(f(x) = x-1\). The derivative is:

\(f'(x) = \frac{d}{dx}(x-1) = 1\)

Step 5: Evaluate Left-Hand and Right-Hand Derivatives at the Point

The left-hand derivative at \(x=1\) is the limit of \(f'(x)\) as \(x\) approaches 1 from the left:

\(f'(1^-) = \lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} (2ax+a)\)

Substitute \(x=1\):

\(2a(1)+a = 3a\)

The right-hand derivative at \(x=1\) is the limit of \(f'(x)\) as \(x\) approaches 1 from the right:

\(f'(1^+) = \lim_{x \to 1^+} f'(x) = \lim_{x \to 1^+} 1\)

This limit is simply:

\(1\)

Step 6: Set up the Differentiability Equation

For the function to be differentiable at \(x=1\), the left-hand derivative must equal the right-hand derivative:

\(f'(1^-) = f'(1^+)\)

\(3a = 1 \quad (Equation\;2)\)

Step 7: Solve the System of Equations for a and b

We now have a system of two equations with two variables \(a\) and \(b\):

• Equation 1 (from continuity): \(2a+b = 0\)
• Equation 2 (from differentiability): \(3a = 1\)

From Equation 2, we can directly find the value of \(a\):

\(3a = 1 \implies a = \frac{1}{3}\)

Now substitute the value of \(a\) into Equation 1 to find the value of \(b\):

\(2a+b = 0\)

\(2\left(\frac{1}{3}\right) + b = 0\)

\(\frac{2}{3} + b = 0\)

\(b = -\frac{2}{3}\)

Step 8: Find the Value of (a + b)

Finally, we need to find the value of \(a+b\):

\(a+b = \frac{1}{3} + \left(-\frac{2}{3}\right)\)

\(a+b = \frac{1-2}{3}\)

\(a+b = -\frac{1}{3}\)

Summary of Finding (a + b)

By applying the conditions for differentiability (continuity and equality of left/right derivatives) at \(x=1\), we found the values of \(a\) and \(b\) and subsequently calculated their sum.

The value of \(a+b\) is \(-\frac{1}{3}\).

Revision Table: Key Concepts

These conditions are fundamental when dealing with piecewise functions and their smoothness at the points where their definition changes.

Additional Information on Continuity and Differentiability

Continuity: A function is continuous at a point if its graph can be drawn without lifting the pen. Formally, it means the function is defined at that point, the limit exists at that point, and the limit equals the function value.

Differentiability: A function is differentiable at a point if it has a well-defined tangent line at that point. This means the slope of the tangent line (the derivative) exists and is the same whether you approach the point from the left or the right. If a function is differentiable at a point, it must be continuous at that point. However, the reverse is not true; a function can be continuous at a point but not differentiable (e.g., the function \(f(x) = |x|\) at \(x=0\)).

For piecewise functions, continuity and differentiability checks are crucial at the transition points where the function's definition changes, like \(x=1\) in this problem.