Problem 59 - Thomas Calculus

Question

Graph \(y=1 /(2 \sqrt{x})\) in a window that has \(0 \leq x \leq 2 .\) Then, on the same screen, graph $$ y=\frac{\sqrt{x+h}-\sqrt{x}}{h} $$ for \(h=1,0.5,0.1 .\) Then try \(h=-1,-0.5,-0.1 .\) Explain what is going on.

Step-by-Step Solution

Answer

As \( h \) approaches zero in both cases, the graphs approximate the derivative of \( \sqrt{x} \).

Step 1: Understand the Function

We need to graph the function \( y = \frac{1}{2\sqrt{x}} \) which is a reciprocal function with a square root in the denominator. This suggests the graph will be undefined at \( x=0 \) and will approach zero as \( x \) increases.

Step 2: Define the Graphing Window

The graphing window is defined with \( 0 \leq x \leq 2 \). This means that the x-axis will start at 0 and end at 2, allowing us to see how the function behaves over this interval.

Step 3: Graph the Main Function

Plot the graph of \( y = \frac{1}{2\sqrt{x}} \) in the defined window. It should start from near the y-axis at x approaching zero and then gradually curve down towards the x-axis as x approaches 2.

Step 4: Understanding the Second Function

The second function is \( y = \frac{\sqrt{x+h}-\sqrt{x}}{h} \), representing the difference quotient of the square root function. This is connected to the derivative of \( \sqrt{x} \).

Step 5: Graph the Second Function for Positive h

Plot the graph of \( y = \frac{\sqrt{x+h}-\sqrt{x}}{h} \) for \( h=1, 0.5, 0.1 \). You will see the curves get closer to the derivative of \( \sqrt{x} \) as \( h \) decreases, demonstrating the tangent slope at various points.

Step 6: Graph the Second Function for Negative h

Now graph \( y = \frac{\sqrt{x+h}-\sqrt{x}}{h} \) for \( h = -1, -0.5, -0.1 \). Notice how the curves represent a backward difference quotient, also approaching the derivative of \( \sqrt{x} \) as \( h \) becomes smaller negatively.

Step 7: Analyze the Behavior of the Graphs

The sequence of graphs for different \( h \) values, both positive and negative, all approach the same graph: the derivative of \( \sqrt{x} \). This shows how the slopes of secant lines between points on \( \sqrt{x} \) approximate the tangent (derivative) as \( h \) approaches zero.

Key Concepts

Reciprocal Function

The concept of a reciprocal function involves flipping the value of a given input. For a function of the form \( y = \frac{1}{x} \), the reciprocal function is defined as the multiplicative inverse. Rather than producing values that grow indefinitely as \( x \) increases, it yields values that decrease towards zero, reflecting the nature of division by increasingly larger numbers. In our specific exercise, we deal with the reciprocal of a square root function: \( y = \frac{1}{2\sqrt{x}} \). This means at \( x = 0 \), the function is undefined since dividing by zero is impossible.

1. As \( x \) approaches zero from the right, \( \sqrt{x} \) becomes very small, leading the reciprocal to grow larger.
2. As \( x \) increases beyond zero, the value of \( 2\sqrt{x} \) increases, causing \( \frac{1}{2\sqrt{x}} \) to approach zero.

Understanding these behaviors is crucial for visualizing the function's graph which starts from high values near \( x = 0 \) and descends towards the horizontal axis as \( x \) gains value.

Square Root Function

A square root function, defined as \( y = \sqrt{x} \), is characterized by the property that it takes the positive square root of the input value \( x \). This function grows slower compared to linear or quadratic functions because it increases at a decreasing rate. In the exercise context, the square root function helps us explore the difference quotient of a function, a key idea for understanding derivatives.

• The graph of \( y = \sqrt{x} \) is a curve that starts at the origin (0,0) and gently rises, making it an increasing, but concave down, function.
• This gentle rise highlights how each sequential increase in \( x \) contributes less and less to the total increase of \( y \), reflective of square roots scaling in magnitude.

The slower growth of this function sets a foundation for examining changes over small intervals, which we delve into with the difference quotient.

Difference Quotient

The difference quotient is a fundamental tool in calculus used to measure how a function changes as the input changes. It's declared generally as \( \frac{f(x+h) - f(x)}{h} \). In the exercise provided, we use \( y = \frac{\sqrt{x+h} - \sqrt{x}}{h} \), specifically to explore the rate of change of the square root function over a defined interval.

Purpose and Utility:

• The difference quotient measures the average rate of change of \( \sqrt{x} \) over a small interval \( h \).
• It acts as a precursor to understanding the derivative, showcasing how close the secant line's slope comes to the tangent line as \( h \) heads towards zero.

As \( h \) decreases, either positively or negatively, the graphical representation well approximates the tangent line at any point on \( \sqrt{x} \). Thus, for very small \( h \), the difference quotient nearly equates to the actual derivative value.

Derivative of a Function

Derivatives represent the core of calculus. They indicate the rate at which a function is changing at any given point, acting as a measure of the slope of the tangent line to the function's graph at that point. For \( \sqrt{x} \), the derivative can be found through the limit definition, essentially converging from the difference quotient. Mathematically, when \( h \to 0 \), \( \frac{\sqrt{x+h} - \sqrt{x}}{h} \) approaches the derivative of \( \sqrt{x} \), which is \( \frac{1}{2\sqrt{x}} \).

Understanding Derivatives through Difference Quotients:

• As we shrink \( h \), the appearance of the secant turns into that of a tangent.
• The derivative is foundational for discussing concepts like slope and approximation, connecting dynamic physical changes back to mathematical representation.

In practical terms, calculating \( \frac{1}{2\sqrt{x}} \) using derivative approaches provides analytic strength, revealing the instantaneous rate of increase of \( \sqrt{x} \) as \( x \) transitions through continuous values.