Problem 19 - The Calculus with Analytic Geometry

Question

Find the values of \(a\) and \(b\) so that \(f^{\prime}(1)\) exists if $$ f(x)= \begin{cases}x^{2} & \text { if } x<1 \\ a x+b & \text { if } x \geq 1\end{cases} $$

Step-by-Step Solution

Answer

The values are \(a = 2\) and \(b = -1\).

Step 1: Understand Function Definition

The function is defined piecewise. For values of x less than 1, the function is defined as \(f(x) = x^2\). For values of x greater than or equal to 1, the function is defined as \(f(x) = ax + b\). To ensure that \(f^{\tiny{\text{prime}}}(1)\) exists, we must make sure the function is continuous and differentiable at \(x = 1\).

Step 2: Ensure Continuity at \(x = 1\)

First, check the continuity at \(x = 1\) by making sure the limit of \(f(x)\) as \(x\) approaches 1 from the left equals the limit as \(x\) approaches 1 from the right, which also must equal \(f(1)\). For \(x < 1\), \(\lim_{x \to 1^{-}} f(x) = 1^2 = 1\). For \(x \to 1^{+}\), \(\lim_{x \to 1^{+}} f(x) = a(1) + b = a + b\). To ensure continuity, these must be equal: \(1 = a + b\).

Step 3: Differentiate the Function

Differentiate both parts of the function. For \(x<1\), the derivative \(f'(x) = 2x\). For \(x \geq 1\), the derivative \(f'(x) = a\).

Step 4: Ensure Differentiability at \(x = 1\)

Check if the derivative from the left equals the derivative from the right at \(x = 1\) for differentiability. For \(x<1\), \(f'(1^{-}) = 2(1) = 2\). For \(x \geq 1\), \(f'(1^{+}) = a\). Thus, \(a = 2\).

Step 5: Solve for \(b\)

Use the continuity equation \(1 = a + b\) and substitute \(a = 2\): \(1 = 2 + b\), which simplifies to \(b = -1\).

Key Concepts

Continuity

To understand the concept of continuity, let's start with its definition. A function is said to be continuous at a point if the limit of the function as it approaches that point from both sides is equal to the function's value at that point. In mathematical terms, for a function \( f(x) \) to be continuous at \( x = c \), it must satisfy the following condition: \( \text{lim}_{x \to c^-} f(x) = \text{lim}_{x \to c^+} f(x) = f(c) \). For the given piecewise function, continuity at \( x = 1 \) means that the limit from the left, \(\text{lim}_{x \to 1^-} f(x) \), should equal the limit from the right, \(\text{lim}_{x \to 1^+} f(x) \), and also equal \( f(1) \). In our case, \( f(x) \) for \( x < 1 \) is \( x^2 \), so \(\text{lim}_{x \to 1^-} f(x) = 1 \). For \( x \geq 1 \), \( f(x) = ax + b \), so \(\text{lim}_{x \to 1^+} f(x) = a + b \). To have continuity at \( x = 1 \), we must have: \( 1 = a + b \). This condition ensures that the function does not break at the point of transition.

Differentiability

Differentiability is a stronger condition than continuity. A function is differentiable at a point if it has a derivative at that point, meaning the function's gradient or slope changes smoothly without any sudden jumps or breaks. For a function \( f(x) \) to be differentiable at \( x = c \), the derivative from the left must equal the derivative from the right: \( \text{lim}_{x \to c^-} f'(x) = \text{lim}_{x \to c^+} f'(x) \). For the piecewise function given, we need the derivative for \( x < 1 \) and for \( x \geq 1 \) to be equal at \( x = 1 \). First, we find the derivatives: for \( x < 1 \), \( f'(x) = 2x \); for \( x \geq 1 \), \( f'(x) = a \). At \( x = 1 \), \( \text{lim}_{x \to 1^-} f'(x) = 2 \times 1 = 2 \) and \( \text{lim}_{x \to 1^+} f'(x) = a \). To ensure differentiability, \( 2 = a \). This guarantees that the function's slope is consistent at the transition point.

Derivatives

Derivatives measure the rate at which a function is changing at any given point, effectively providing the slope of the function's graph at that point. For a function \( f(x) \), the derivative, noted as \( f'(x) \), is the limit: \( f'(x) = \text{lim}_{h \to 0} \frac{f(x+h) - f(x)}{h} \). In simpler terms, it tells you how much \( f(x) \) changes as \( x \) changes. For the given piecewise function: - For \( x < 1 \), the function is \( x^2 \), and its derivative is \( 2x \). - For \( x \geq 1 \), the function is \( ax + b \), and its derivative is \( a \). To ensure a smooth transition at \( x = 1 \), these derivatives must match as we calculated earlier, requiring \( a = 2 \). Knowing derivatives helps in understanding the behavior and the rate of change of the function at different points.

Piecewise Functions

A piecewise function is a function that is defined by different expressions for different intervals of the domain. Understanding piecewise functions is crucial because they represent real-world situations where a single rule might not apply to all circumstances. For the given function, we have: \( f(x)= \begin{cases}x^2 & \text { if } x<1 \ a x + b & \text { if } x \text { greater than or equal to }1\text. \). Analyzing such functions involves checking: - Continuity at the transition points - Differentiability at the transition points By ensuring both of these, we can find values for \( a \) and \( b \) that make the function behave smoothly across its entire range. Piecewise functions help model complex behaviors that can't be captured by a single uniform formula, making them essential in higher mathematics and applied fields.