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Finding the Slope of a Graph In Exercises \(63-70\) , find the slope of the graph of the function at the given point. Use the derivative feature of a graphing utility to confirm your results. $$y=\frac{4}{(x+2)^{2}}, \quad(0,1)$$

Short Answer

Expert verified
The slope of the graph of the given function at the point (0,1) is -1.

Step by step solution

01

Find the Derivative of the Function

Consider the given function \(y=\frac{4}{(x+2)^{2}}\). The derivative of \(y\) with respect to \(x\) (denoted as \(y'\) or \(dy/dx\)) is calculated using the chain rule for derivatives and the power rule for \(n=-2\). Here, you can treat \((x+2)^{2}\) as 'u'.So, \(y'= -2\cdot \frac{4}{(x+2)^{3}} = -\frac{8}{(x+2)^{3}} \).
02

Evaluate the Derivative at the Given Point

Next, to find the slope of the function at the given point \((0,1)\), substitute \(x=0\) in \(y'\), the derivative. The result gives the slope of the function at \((0,1)\).So, if \(y'=-\frac{8}{(x+2)^{3}}\) then substituting \(x=0\) gives slope \(y'=-\frac{8}{8} = -1\).
03

Confirmation Using Graphing Utility

For validation, enter the function and derivative function into a graphing utility. Plot both graphs and visually inspect if the slope of the tangent to the curve at \((0,1)\) is -1. Also, check for an instantaneous rate of change at \((0,1)\) in the derivative graph. The slope found and graphical illustration should match.

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Key Concepts

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

Derivatives
Derivatives are a fundamental concept in calculus that measures how a function changes as its input changes. When we find the derivative of a function, we effectively find the instantaneous rate of change of the function at any point. This is incredibly useful in many real-world applications, such as physics, engineering, and economics. The notation for derivatives can vary; you might see it represented as \( f'(x) \), \( rac{dy}{dx} \), or \( Df \).
  • The derivative provides the slope of the tangent line to the curve at any given point.
  • The process of finding the derivative is called differentiation.

In solving problems, it's important to understand the basic rules of differentiation, such as the power rule, the product rule, and the chain rule, which help in breaking down complex functions into manageable parts. Differentiation can also help us in optimizing functions, understanding motion, and predicting future behavior of variables in a function.
Slope of a Function
The slope of a function at a particular point informs us about the direction and steepness of a curve. For linear functions, the slope is constant and represents the steepness of the line. However, for nonlinear functions, the slope varies at different points on the curve.
  • The slope is equivalent to the value of the derivative of the function at a particular point.
  • It can tell us if the function is increasing, decreasing, or remaining constant at that point.

In the original exercise, we found that the slope of the function \( y = \frac{4}{(x+2)^2} \) at the point (0,1) is -1. This tells us that at this point, the curve is decreasing. The negative slope means that as \( x \) increases slightly from 0, \( y \) decreases. Understanding slope helps in many fields, such as economics, where it might represent costs or revenues increasing or decreasing over time.
Chain Rule
The chain rule is a powerful tool in calculus used to find the derivative of a composite function. When functions are nested within each other, like \( y = \frac{4}{(x+2)^2} \), the chain rule allows us to differentiate them efficiently. For a composite function \( f(g(x)) \), the chain rule states that the derivative is the derivative of the outer function \( f \) evaluated at \( g(x) \), multiplied by the derivative of the inner function \( g(x) \).
  • In mathematical terms, if \( h(x) = f(g(x)) \), then \( h'(x) = f'(g(x)) \cdot g'(x) \).
  • This rule helps simplify the process of taking derivatives of complex functions quickly.

In our exercise, we treated \( (x+2)^2 \) as \( u \), allowing us to apply the chain rule followed by the power rule to find the derivative. This approach provides a systematic way to handle more complicated functions.
Graphing Utility
A graphing utility is a technological tool or software that helps visualize mathematical equations and functions. Graphing utilities often have features that facilitate calculus operations like differentiation and integration. These tools allow us to confirm results obtained by analytical methods and offer a visual representation of functions.
  • Graphing utilities can be standalone devices, apps, or online platforms.
  • They help by providing an intuitive 'picture' of complex mathematical concepts.

In the original exercise, we used a graphing utility to verify the calculated slope at the point \( (0,1) \). By inputting the original function and its derivative, the graphing utility can display the function, its tangent line, and demonstrate that the tangent's slope at \( (0,1) \) is indeed -1. This visual confirmation is an invaluable step in ensuring that analytical work aligns with graphical representations.

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Most popular questions from this chapter

True or False? In Exercises \(121-124,\) determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. The slope of the function \(f(x)=\sin a x\) at the origin is \(a\)

Graphical Analysis In Exercises 81-84, use a graphing utility to graph the function and find the x-values at which f is differentiable. $$f(x)=|x-5|$$

FOR FURTHER INFORMATION For more information on the mathematics of moving ladders, see the article "The Falling Ladder Paradox" by Paul Scholten and Andrew Simoson in The College Mathematics Journal. To view this article, go to MathArticles.com. Construction A construction worker pulls a five-meter plank up the side of a building under construction by means of a rope tied to one end of the plank (see figure). Assume the opposite end of the plank follows a path perpendicular to the wall of the building and the worker pulls the rope at a rate of 0.15 meter per second. How fast is the end of the plank sliding along the ground when it is 2.5 meters from the wall of the building?

Linear vs. Angular Speed A patrol car is parked 50 feet from a long warehouse (see figure). The revolving light on top of the car turns at a rate of 30 revolutions per minute. How fast is the light beam moving along the wall when the beam makes angles of (a) \(\theta=30^{\circ},\) (b) \(\theta=60^{\circ},\) and (c) \(\theta=70^{\circ}\) with the perpendicular line from the light to the wall?

Electrical Circuit The voltage \(V\) in volts of an electrical circuit is \(V=I R,\) where \(R\) is the resistance in ohms and I is the current in amperes. \(R\) is increasing at a rate of 2 ohms per second, and \(V\) is increasing at a rate of 3 volts per second. At what rate is \(I\) changing when \(V=12\) volts and \(R=4\) ohms?

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