/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 a) Graph the function. b) Draw... [FREE SOLUTION] | 91Ó°ÊÓ

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a) Graph the function. b) Draw tangent lines to the graph at points whose \(x\) -coordinates are \(-2,0,\) and 1 c) Find \(f^{\prime}(x)\) by determining \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\). d) Find \(f^{\prime}(-2), f^{\prime}(0),\) and \(f^{\prime}(1) .\) These slopes should match those of the lines you drew in part (b). $$f(x)=-3 x^{2}$$

Short Answer

Expert verified
The slopes of the tangent lines at \( x = -2, 0, \) and \( 1 \) are \( 12, 0, \) and \( -6 \) respectively.

Step by step solution

01

Graph the Function

Plot the function \( f(x) = -3x^2 \) on a coordinate plane. This function is a downward-opening parabola, with its vertex at the origin \((0,0)\). The graph is symmetrical about the y-axis.
02

Draw Tangent Lines

To draw tangent lines at the points where \( x = -2, 0, \) and \( 1 \), determine their corresponding y-values on the graph. At \( x = -2 \), \( y = -3(-2)^2 = -12 \). At \( x = 0 \), \( y = -3(0)^2 = 0 \). At \( x = 1 \), \( y = -3(1)^2 = -3 \). Draw tangent lines that touch the graph at these points.
03

Calculate the Derivative

Find the derivative \( f'(x) \) using the limit definition: \ \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]. \ Substitute \( f(x) = -3x^2 \) giving: \ \[ f'(x) = \lim_{h \to 0} \frac{-3(x+h)^2 + 3x^2}{h} \]. \ Simplify: \ \[ f'(x) = \lim_{h \to 0} \frac{-3(x^2 + 2xh + h^2) + 3x^2}{h} = \lim_{h \to 0} \frac{-3x^2 - 6xh - 3h^2 + 3x^2}{h} \]. \ Simplifying further, \ \[ f'(x) = \lim_{h \to 0} \frac{-6xh - 3h^2}{h} = \lim_{h \to 0} (-6x - 3h) \]. \ Conclude: \[ f'(x) = -6x \].
04

Find Derivatives at Specific Points

Evaluate \( f'(x) \) for \( x = -2, 0, \) and \( 1 \). First, for \( x = -2 \), \( f'(-2) = -6(-2) = 12 \). Next, for \( x = 0 \), \( f'(0) = -6(0) = 0 \). Finally, for \( x = 1 \), \( f'(1) = -6(1) = -6 \). Compare these slopes to the tangent lines drawn in Step 2.

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Key Concepts

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

Derivative
The derivative is a fundamental tool in calculus used to determine how a function changes at any given point. In essence, it tells you the "rate of change" or "slope" of the function. For example, consider the function \( f(x) = -3x^2 \). The derivative, represented as \( f'(x) \), gives us the slope of the tangent line to the graph of the function at any point \( x \).Key points about derivatives:
  • They show how fast or slow a function is changing.
  • Positive derivatives indicate an increasing function, while negative ones point to a decreasing function.
  • For constant functions, the derivative is zero since there is no change.
In practice, finding the derivative involves calculus techniques such as the limit definition, which lets you calculate the derivative at a point by considering how the function behaves as you make infinitesimally small changes to \( x \).
Limit Definition
The limit definition of the derivative is a crucial concept in calculus. It allows us to compute the derivative of a function at a specific point. The basic idea is to find the slope of the tangent line to the curve at that point.Mathematically, the limit definition is expressed as:\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]This expression means you take two points close to each other on the curve \( f(x) \) and look at the slope between them as the distance \( h \) approaches zero.This process involves:
  • Choosing a small \( h \) value
  • Calculating \( f(x+h) \) and \( f(x) \)
  • Subtracting \( f(x) \) from \( f(x+h) \)
  • Dividing by \( h \) and taking the limit as \( h \) goes to zero
In our example with \( f(x) = -3x^2 \), using the limit definition gives us a derivative \( f'(x) = -6x \). This shows that the change in \( f \) is proportional to \( x \), scaled by \(-6\).
Tangent Lines
Tangent lines play an essential role in understanding how functions behave. A tangent line to a graph at a given point is a straight line that "just touches" the curve at that point. It represents the instantaneous rate of change of the function, or the derivative, at that specific point.Here's how to find tangent lines at certain points:
  • Identify the point where you want the tangent.
  • Compute the derivative at that point, which provides the slope of the tangent line.
  • Use the point-slope form of a line \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the point of tangency.
For the function \( f(x) = -3x^2 \), we drew tangent lines at \( x = -2, 0, \) and \( 1\). Calculated slopes (12, 0, and -6 respectively) match the graph, verifying the validity of both our derivative and visual checks.
Parabola
A parabola is a U-shaped curve that can open upwards or downwards. It is the graph of a quadratic function, such as \( f(x) = ax^2 + bx + c \). In our example, the function \( f(x) = -3x^2 \) describes a downward-opening parabola, with its vertex at the origin \((0,0)\).Characteristics of a parabola:
  • The vertex is the highest or lowest point of the curve, depending on its orientation.
  • If \( a > 0 \), the parabola opens upwards; if \( a < 0 \), it opens downwards.
  • It is symmetrical about its line of symmetry, often the y-axis if the function is simple.
Understanding parabolas is crucial since they often appear in physics and engineering contexts when analyzing patterns of motion and forces. In calculus, their simple derivatives and clear graphical characteristics make them excellent examples to study tangents and dynamic changes.

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Most popular questions from this chapter

For each function, find the points on the graph at which the tangent line has slope 1 . $$ y=\frac{1}{3} x^{3}-x^{2}-4 x+1 $$

It has been shown that the home range, in hectares, of a carnivorous mammal weighing \(w\) grams can be approximated by $$ H(w)=0.11 w^{1.36} $$ (Source: Based on information in Emlen, J. M., Ecology: An Evolutionary Approach, p. \(216,\) Reading, MA: Addison-Wesley, 1973 ; and Harestad, A. S., and Bunnel, F. L., "Home Range and Body Weight-A Reevaluation," Ecology, Vol. 60, No. 2, pp. 405-418.) a) Find the average rate at which a carnivorous mammal's home range increases as the animal's weight grows from \(500 \mathrm{~g}\) to \(700 \mathrm{~g}\). b) Find \(\frac{H(300)-H(200)}{300-200}\). What does this rate represent?

Find dy/dx. Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$ y=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} $$

At the beginning of a trip, the odometer on a car reads \(30,680,\) and the car has a full tank of gas. At the end of the trip, the odometer reads \(31,077 .\) It takes 13.5 gal of gas to refill the tank. a) What is the average rate at which the car was traveling, in miles per gallon? b) What is the average rate of gas consumption in gallons per mile?

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'Hôpital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow-3}\left(\frac{x^{2}-9}{x+3}\right) $$

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