/*! 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 15 Let \(f(x)=2 x^{2}-5 x-1 .\) Let... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(f(x)=2 x^{2}-5 x-1 .\) Let \(P\) and \(Q\) be points on the graph of \(f\) with \(x\) -coordinates 3 and \(3+h\), respectively. (a) Sketch the graph of \(f\) and the secant lines through \(P\) and \(Q\) for \(h=1\) and \(h=0.1\). (b) Find the slope of the secant line through \(P\) and \(Q\) for \(h=1, h=0.1\), and \(h=0.01\). (c) Find the slope of the tangent line to \(f\) at point \(P\) by calculating the appropriate limit. (d) Find the equation of the line tangent to \(f\) at point \(P\).

Short Answer

Expert verified
The slopes of the secant lines for \(h=1\), \(h=0.1\), and \(h=0.01\) can be calculated by substituting these \(h\) values into \((f(x+h)-f(x))/h\). The slope of the tangent line at point \(P\) is the limit as \(h\) approaches \(0\) of the same expression. Once the slope is known, the equation of the tangent line can be formulated using the point-slope form.

Step by step solution

01

Graph of function and secant lines

Use a graphing tool to sketch the graph of \(f(x)=2x^{2} - 5x - 1\). To sketch the secant lines through \(P\) and \(Q\) for \(h=1\) and \(h=0.1\), first calculate the \(y\)-coordinates for points \(P\) and \(Q\) using the function \(f\). Then, visually construct the secant lines on the graph.
02

Slope of Secant Lines

The slope of the secant line through points \(P\) and \(Q\) is given by the difference formula \((f(x+h)-f(x))/h\). Substitute \(x=3\) and \(h=1,\,0.1,\,0.01\) into the formula to get the slopes for each \(h\).
03

Slope of Tangent Line

To find the slope of the tangent line at point \(P\), we need to take the limit as \(h\) approaches \(0\) of \((f(x+h)-f(x))/h\). Substituting \(x=3\) into the limit gives us the required slope.
04

Equation of Tangent Line

The equation of the tangent line, using point-slope form, is \(y - y_{1} = m(x - x_{1})\). Here, \(x_{1}\) and \(y_{1}\) are the coordinates of point \(P\), which would be \(3\) and \(f(3)\) respectively, and \(m\) is the slope of the tangent line as calculated in the previous step.

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Key Concepts

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

Tangent Line Slope
Understanding the slope of a tangent line is crucial in calculus, especially when discussing the instantaneous rate of change at a point. To find the slope of the tangent line at a point on a curve, we use the concept of a limit, which leads us to the derivative of the function at that point.

As illustrated in the exercise, you calculate the slope of the secant line by taking the difference in the y-values of two points on the function and dividing it by the difference in their x-values. This gives us an average rate of change between those two points. To get the slope of the tangent line, which is the instantaneous rate of change at a single point, we consider the secant line as the two points get infinitely close together — this is where the limiting process comes in.

Mathematically, the slope of the tangent line at point P, if P has coordinates \(x, f(x)\), is found by evaluating the derivative at that point, which is the limit of the slope of the secant lines as the distance \(h\) approaches zero: \[ \lim_{{h \to 0}}\frac{{f(x+h)-f(x)}}{h} \].

This fundamental process is key to understanding many concepts in calculus and physics, such as velocity, acceleration, and other rates of change.
Graphing Quadratic Functions
A quadratic function can be identified by its general form \(f(x) = ax^2 + bx + c\), where \(a, b,\) and \(c\) are constants, and \(a \eq 0\). Graphing such functions is a foundational skill in algebra and precalculus.

The graph of a quadratic function is a parabola, which can either open upwards or downwards depending on the sign of the coefficient \(a\). To graph the quadratic function given in the exercise, \(f(x) = 2x^2 - 5x - 1\), you would plot the vertex, which is the highest or lowest point on the graph, and then use additional points to define the shape.

The axis of symmetry can be found using the formula \(x = -\frac{b}{2a}\), and the vertex can be found by plugging the x-coordinate of the axis of symmetry back into the function. Then, plot the y-intercept (where \(x=0\)) and other points by choosing suitable x-values and finding their corresponding y-values. Connect these points to form the parabola.
Limit Definition of a Derivative
The limit definition of a derivative is at the heart of calculus. It enables us to find the derivative of a function, which represents the slope of the tangent line to the function at any point. As seen in the exercise for finding the slope of a tangent line, the derivative is defined as the limit of the slopes of the secant lines as they approach the point of tangency.

The formal limit definition of the derivative of function \(f\) at any point \(x\) is: \[ f'(x) = \lim_{{h \to 0}} \frac{{f(x + h) - f(x)}}{h} \].

In practice, this means that to find the derivative of a function, we calculate the limit as the secant lines get closer and closer to being the tangent line. By setting \(h\) to smaller and smaller values, we approximate this process until we can define the exact slope at a single point when \(h\) is infinitesimally close to zero.
Equation of a Tangent Line
Once the slope of the tangent line at a point on a function is known, the next step is to write the equation of this line. This allows us to create a linear function that just touches the curve at that point and has the same slope as the curve does at that point.

The equation of a tangent line to a function \(f\) at a point \(P\) with coordinates \(x_1, f(x_1)\) is given by the point-slope form of a line: \[ y - f(x_1) = m(x - x_1) \], where \(m\) is the slope of the tangent line, which is the derivative of \(f\) at \(x_1\).

In the exercise, once you've calculated the slope \(m\) using the limit definition of a derivative, you plug in the coordinates of point \(P\) and the slope into the point-slope formula to find the equation of the tangent line. This linear equation allows us to closely examine the behavior of the function at the point of tangency and is a powerful tool in both theoretical and applied mathematics.

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Most popular questions from this chapter

The graph of \(g(x)=x^{2}-4\) looks just like the graph of \(f(x)=x^{2}\) shifted vertically downward 4 units. Will \(f^{\prime}\) and \(g^{\prime}\) be the same or different? Explain your reasoning.

Let \(f(x)=\frac{3}{x-5}\). (a) Using the limit de nition of derivative, nd \(f^{\prime}(2)\). (b) Find two ways of checking whether or not your answer is reasonable. These methods should not involve simply checking your algebra. They can be numerical or graphical use your ingenuity.

Let \(g\) be a function that is locally linear. We know that \(g^{\prime}(a)\) is the slope of the tangent line to the graph of \(g\) at point \(A=(a, g(a))\). Let \(Q=(t, g(t))\) be an arbitrary point on the graph of \(g, Q\) distinct from \(A\). (a) Write a difference quotient (i.e., an expression of the form \(\frac{?-?}{2-?}\), the quotient of two differences) that gives the slope of the secant line through points \(A\) and \(Q\). (b) Take the appropriate limit of the difference quotient in part (a) to arrive at an expression for \(g^{\prime}(a)\).

A baked apple is taken out of the oven and put into the refrigerator. The refrigerator is kept at a constant temperature. Newton's Law of Cooling says that the difference between the temperature of the apple and the temperature of the refrigerator decreases at a rate proportional to itself. That is, the apple cools down most rapidly at the outset of its stay in the refrigerator, and cools increasingly slowly as time goes by. You have the following pieces of information: At the moment the apple is put in the refrigerator its temperature is 110 degrees and is dropping at a rate of 4 degrees per minute. Twenty minutes later the temperature of the apple is 70 degrees. (a) Let \(T\) be the temperature of the apple at time \(t\), where \(t\) is measured in minutes and \(t=0\) is when the apple is put in the refrigerator. Express the three bits of information provided above in functional notation. Sketch a graph of \(T\) versus \(t\). (b) Using the same set of axes as you did in part (a), draw the cooling curve the baked apple would have if it were cooling linearly, with the initial temperature of 110 degrees and initial rate of cooling of 4 degrees per minute. What would the temperature of the apple be after 15 minutes using this linear model? In reality, is the temperature more or less? (c) Since the apple's temperature dropped from 110 degrees to 70 degrees in twenty minutes, the average rate of change of temperature over the first twenty minutes is \(\frac{-40 \text { degrees }}{20 \text { minutes }}\) or \(-2 \frac{\text { degrees }}{\text { minute }} .\) Using the same set of axes as you did in parts (a) and (b), draw the cooling curve the baked apple would have if it were cooling linearly, with the initial temperature of 110 degrees and rate of cooling of 2 degrees per minute. What would the temperature of the apple be after 15 minutes using this linear model? In reality, is the temperature more or less?

Find the equation of the line tangent to \(y=\frac{2}{x+1}\) at the point \((1,1)\).

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