Problem 68 - Calculus for Biology and Medicine

Question

Graph each function and, on the basis of the graph, guess where the function is not differentiable. (Assume the largest possible domain.) $$ \text { 8. } f(x)=\left\\{\begin{array}{cl} x^{2} & \text { for } x \leq-1 \\ 2-x^{2} & \text { for } x>-1 \end{array}\right. $$

Step-by-Step Solution

Answer

The function is not differentiable at \( x = -1 \).

Step 1: Understand the Function Piecewise Definition

The function \( f(x) \) is given in two pieces. For \( x \leq -1 \), the function is \( f(x) = x^2 \). For \( x > -1 \), the function changes to \( f(x) = 2 - x^2 \). We need to look at each piece separately and understand their behaviors.

Step 2: Analyze Each Piece for Graph Characteristics

The graph of \( x^2 \) is a parabola opening upwards, mostly relevant for \( x \leq -1 \). The graph of \( 2 - x^2 \) is a parabola opening downwards, relevant for \( x > -1 \). Observe how these two parabolas behave at \( x = -1 \), which is the critical point.

Step 3: Check Continuity at x = -1

For continuity at \( x = -1 \), compute \( f(-1) \) for both pieces: \( (-1)^2 = 1 \) and \( 2 - (-1)^2 = 1 \). Both values agree, so \( f(x) \) is continuous at \( x = -1 \).

Step 4: Determine Differentiability at x = -1

To check differentiability at \( x = -1 \), compute the derivatives of each piece at \( x = -1 \). The derivative of \( x^2 \) is \( 2x \), which at \( x = -1 \) gives \(-2\). The derivative of \( 2-x^2 \) is \(-2x \), which at \( x = -1 \) also gives \(2\). Since \(-2 eq 2\), \( f(x) \) is not differentiable at \( x = -1 \).

Step 5: Sketch the Graph

Draw the graph for each piece on the same set of axes. For \( x \leq -1 \), sketch \( x^2 \), and for \( x > -1 \), sketch \( 2 - x^2 \). At \( x = -1 \), both segments connect, showing continuity but not differentiability due to differing slopes.

Key Concepts

Differentiability

Differentiability is a key concept in calculus that tells us whether or not a function has a derivative at a particular point. In simpler terms, it indicates whether a function can be "smoothly drawn" or has a well-defined slope at that point.

For a function to be differentiable at a point, the function must be continuous there, and the derivative must exist. This means that there can't be any sharp corners, cusps, or vertical tangents at that point, as these features prevent a derivative from being calculated.

In the context of our exercise, we examine the differentiability of \( f(x) = \begin{cases} x^2 & \text{if } x \leq -1 \ 2 - x^2 & \text{if } x > -1 \end{cases} \) at \( x = -1 \). Though the function is continuous at \( x = -1 \), it is not differentiable because the slopes of the two pieces do not match. Therefore,

• The derivative from the left, \( (-2) \), does not equal the derivative from the right, \( 2 \).
• A difference in slopes results in a sharp corner at \( x = -1 \).

Continuity

Continuity is a fundamental property of functions that make them predictable and easier to work with. When a function is continuous at a particular point, there are no jumps or breaks in the graph of the function at that point.

To determine if a function is continuous at a particular point, the left-hand limit, right-hand limit, and the value of the function at that point all need to be equal. In our piecewise function:

\( f(x) = \begin{cases} x^2 & \text{if } x \leq -1 \ 2 - x^2 & \text{if } x > -1 \end{cases} \),

evaluating continuity at \( x = -1 \) involves comparing these limits.

• The value of the function from the left (using \( x^2 \)) is \( 1 \).
• The value from the right (using \( 2 - x^2 \)) is \( 1 \) as well.
• The actual value of \( f(-1) \) is also \( 1 \).

Since all these are equal, the function is continuous at \( x = -1 \), with no break or jump in the graph.

Parabolas

Parabolas are symmetric, U-shaped curves that are the graphs of quadratic functions of the form \( ax^2 + bx + c \). The direction in which a parabola opens depends on the sign of the coefficient \( a \). If \( a > 0 \), the parabola opens upwards, whereas if \( a < 0 \), it opens downwards.

In our piecewise function:
\( f(x) = \begin{cases} x^2 & \text{for } x \leq -1 \ 2 - x^2 & \text{for } x > -1 \end{cases} \),
we see two parabolas:

• For \( x^2 \) (with \( a = 1 \)), the parabola opens upwards.
• For \( 2 - x^2 \) (effectively \( -x^2 + 2 \)), the parabola opens downwards due to the negative sign of \( x^2 \).

At the junction \( x = -1 \), these two parabolas meet. Even though they connect to show continuity, they don't share the same slope, causing differentiability to fail at this point. Understanding how parabolas behave helps in smoothly transiting between different pieces of piecewise functions while identifying important characteristics like direction and vertex.