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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.

Short Answer

Expert verified
Graphs of difference quotients with varying \( h \) approximate the derivative of \( \sqrt{x} \), \( y = \frac{1}{2\sqrt{x}} \), as \( h \to 0 \).

Step by step solution

01

Analyze the Function y = 1/(2√x)

First, understand the function \( y = \frac{1}{2\sqrt{x}} \). This function is defined for \( x > 0 \) because \( \sqrt{x} \) is not defined for negative values and \( x = 0 \) leads to division by zero. The graph will take the form of a decreasing curve in the positive \( x \) domain.
02

Graph the Function y = 1/(2√x)

Graph the function \( y = \frac{1}{2\sqrt{x}} \) over the interval \( 0 < x \leq 2 \). As \( x \) increases from 0 to 2, the value of \( y \) decreases, forming a downward-sloping curve from the top left to the bottom right of the graph within this range.
03

Analyze and Graph the Difference Quotient y = (√(x+h) - √x)/h for h = 1, 0.5, 0.1

Consider the expression \( y = \frac{\sqrt{x+h} - \sqrt{x}}{h} \) where \( h \) is positive (1, 0.5, 0.1). This expression represents the average rate of change of the function \( \sqrt{x} \) as \( x \) changes by \( h \). Graph each of these on the same axis with the original function. For positive \( h \), as \( h \) decreases, the graph will approximate the derivative of \( \sqrt{x} \) more closely.
04

Analyze and Graph the Difference Quotient for h = -1, -0.5, -0.1

Now, use negative values for \( h \) (-1, -0.5, -0.1). The expression turns into \( y = \frac{\sqrt{x+h} - \sqrt{x}}{h} \) where \( h \) is negative, signifying step backwards from \( x \). Graph these functions. For negative \( h \), as \( h \) approaches zero, the graph will again approximate the derivative of \( \sqrt{x} \), which is the slope of the tangent to the curve.
05

Interpret the Graphs

The behavior of these graphs can be seen as approximations to the derivative of \( \sqrt{x} \), highlighting how the difference quotient for both forward and reverse steps converges to the derivative \( \frac{1}{2\sqrt{x}} \) as \( h \) approaches 0 from both the positive and negative sides.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference Quotient
The difference quotient is a fundamental concept in calculus used to approximate the slope of the tangent line to a curve at a point. It provides a method to find the derivative of a function, serving as a bridge between algebra and calculus. For a function \( f(x) \), the difference quotient is expressed as:\[ y = \frac{f(x+h) - f(x)}{h} \]where \( h \) is a small increment. This expression calculates the average rate of change of the function as \( x \) changes to \( x+h \). The function used in our exercise is \( \sqrt{x} \), so the difference quotient becomes:\[ y = \frac{\sqrt{x+h} - \sqrt{x}}{h} \]This formula helps approximate the slope at \( x \), aligning closely with the concept of a derivative. As \( h \) gets smaller, the difference quotient provides a more accurate estimate of the derivative.
Derivative Approximation
Derivative approximation using the difference quotient is a core concept in calculus, crucial for understanding the rate at which a function changes. When graphing these approximations, we observe patterns forming as \( h \) varies:- For small positive \( h \) (like 0.1), the function's slope is approximated moving forward along the curve.- The less \( h \) is, the better our approximation to the true derivative is.- As \( h \) approaches zero, the difference quotient converges to the actual derivative.It's important to note that a derivative \( \frac{d}{dx}(\sqrt{x}) \) results in \( \frac{1}{2\sqrt{x}} \), the slope of the tangent line at any point of the function \( \sqrt{x} \). By graphing, students visually link the mathematical expression to its graphical counterpart, observing how the approximation improves with different values of \( h \).
Rate of Change
The rate of change is an essential concept in capturing how quickly a function's outputs change in response to changes in its inputs. With a focus on \( \sqrt{x} \), the difference quotient captures the average rate of change over a small interval \( h \):- For positive \( h \), it measures the forward rate of change.- For negative \( h \), it gauges the backward rate of change.The more \( h \) approaches zero, the closer we measure the instantaneous rate of change, which is the derivative. This shift from an average to an instantaneous measure illustrates how calculus builds on core algebraic principles. The goal is a seamless transition from discrete changes to continuous analysis, deepening students' understanding of the function's behavior.
Graph Interpretation
Graph interpretation is pivotal in applying calculus concepts such as derivative approximation. By observing the graphs of \( y = \frac{\sqrt{x+h} - \sqrt{x}}{h} \) with varying \( h \), students discern the effect of \( h \) on graph shape. - When \( h \) is positive and small, the graph closely mirrors the tangent line, indicating the rate of change at point \( x \).- For negative \( h \), we see a tangential approach from the left, providing similar insights.The critical insight gained is how both forward and backward differences approximate the tangent's slope as \( h \) narrows. This visual representation solidifies the connection between algebraic calculation and graphical trends. Observing these graphs, learners experience firsthand how calculus extrapolates real-world dynamic shifts, ingraining deeper mathematical comprehension.

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Most popular questions from this chapter

Use a CAS to perform the following steps a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\). c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$x y^{3}+\tan (x+y)=1, \quad P\left(\frac{\pi}{4}, 0\right)$$

Find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=u+\frac{1}{\cos ^{2} u}, \quad u=g(x)=\pi x, \quad x=1 / 4$$

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\). Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\) c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=x^{2 / 3}(x-2), \quad[-2,3], \quad a=2$$

Find \(d y / d t\) $$y=\left(\frac{3 t-4}{5 t+2}\right)^{-5}$$

When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity \(g\). The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in \(g.\) By keeping track of \(\Delta T,\) we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L\). a. With \(L\) held constant and \(g\) as the independent variable, calculate \(d T\) and use it to answer parts (b) and (c). b. If \(g\) increases, will \(T\) increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a \(100-\mathrm{cm}\) pendulum is moved from a location where \(g=980 \mathrm{cm} / \mathrm{s}^{2}\) to a new location. This increases the period by \(d T=0.001\) s. Find \(d g\) and estimate the value of \(g\) at the new location.

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