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Chapter 5: Problem 3

Let \(f(x)=x^{2}\). Find the point at which the line tangent to \(f(x)\) at \(x=2\) intersects the line tangent to \(f(x)\) at \(x=-1\).

Short Answer

Expert verified
The point of intersection of the two tangent lines at \(x=2\) and \(x=-1\) to the function \(f(x)=x^{2}\) is \((5/6, -1/3)\).

Step by step solution

01

Find the Derivative of \(f(x)\)

The derivative of a function gives us its slope at any point. Since the slope of a tangent line to a function at a certain point is simply the derivative of the function at that point, we start by finding the derivative of \(f(x)\). The derivative of \(f(x)=x^{2}\) is \(f'(x)=2x\).
02

Find the Equations of the Tangent Lines

The equation of a tangent line to a function at a certain point can be found using the point-slope form of a line, which is \(y-y_{1}=m(x-x_{1})\), where \((x_{1}, y_{1})\) is a point on the line, and \(m\) is the slope of the line. The slope of the tangent line is the derivative of the function at that point. Let's find the equations for the tangent lines at \(x=2\) and \(x=-1\): \n\nFor \(x=2\), the slope \(m=f'(2)=2(2)=4\), and the point on the line is \((2, f(2))=(2, 2^{2})=(2, 4)\). So, the equation of the tangent line at \(x=2\) is \(y-4=4(x-2)\). Simplifying this gives us \(y=4x-4\). \n\nFor \(x=-1\), the slope \(m=f'(-1)=2(-1)=-2\), and the point on the line is \((-1, f(-1))=(-1, (-1)^{2})=(-1, 1)\). So, the equation of the tangent line at \(x=-1\) is \(y-1=-2(x+1)\). Simplifying this gives us \(y=-2x+1\).
03

Find the Intersection Point of the Two Tangent Lines

To find the intersection point of the two tangent lines, we set their equations equal to each other and solve for \(x\). This gives us \(4x-4=-2x+1\). Solving for \(x\) gives \(x=5/6\). Substituting \(x=5/6\) back into either of the equations for the tangent lines gives the corresponding \(y\) coordinate. Let's substitute \(x=5/6\) into \(y=4x-4\). This gives us \(y=4(5/6)-4=-1/3\). So, the intersection point of the two tangent lines is \((5/6, -1/3)\).

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Key Concepts

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

Derivative
In calculus, a derivative is a powerful tool that helps us understand how a function is changing at any given point. Imagine it as a measure of a function's rate of change, or simply, how steep a curve is at a particular spot.
For instance, if we have a function like \( f(x) = x^2 \), the derivative is found using differentiation rules. Here, the derivative \( f'(x) = 2x \) tells us how steep the function is at any point \( x \).
Why is this important?
  • The derivative provides critical information about the function's behavior.
  • It helps in computing the slope of tangent lines which are central to understanding curves.
The derivative is a foundation for several concepts in calculus, including tangent lines, and is crucial for tackling real-world problems involving rates of change.
Tangent Line
A tangent line is a straight line that just touches a curve at a specific point, matching the curve's slope at that point.
If you've ever seen a graph, visualizing a tangent line can be thought of like seeing the best straight line approximation of the curve at that point.
For example, take the function \( f(x) = x^2 \). At \( x = 2 \), the derivative tells us the slope of the function is 4, which dictates the slope of the tangent line there:
  • A tangent line equation uses the point-slope form \( y - y_1 = m(x - x_1) \).
  • At \( x = 2 \), it intersects the function at \( (2, 4) \), forming the line: \( y = 4x - 4 \).
This line looks like it just skims the curve of \( x^2 \), showing it's tangent at \( x = 2 \).
Tangent lines are extremely useful in calculus and physics for linear approximations and understanding properties of curves.
Intersection Point
An intersection point in geometry occurs where two lines meet or cross each other. Finding this point is a common task in calculus, especially for tangent lines.
In our example, where we found tangent lines to the curve \( f(x) = x^2 \) at \( x = 2 \) and \( x = -1 \), we have two lines:
  • \( y = 4x - 4 \) for \( x = 2 \).
  • \( y = -2x + 1 \) for \( x = -1 \).
To find where these lines intersect, we set their equations equal: \( 4x - 4 = -2x + 1 \).
Solving this provides \( x = \frac{5}{6} \).
Substituting back into one of the equations gives \( y = -\frac{1}{3} \).
Thus, the intersection point of the two lines is \( \left( \frac{5}{6}, -\frac{1}{3} \right) \). Intersection points are critical for solving complex problems in calculus and provide insights into relationships between different functions and their tangents.
Slope
Slope is a simple yet key concept in calculus that describes the steepness or incline of a line. This measurement is the same as the derivative of a function at a given point.
When you think of slope, think of it as the rise over run - how much a line goes up for a given distance it goes across.
Calculating slope for places where curves exist is made possible using derivatives:
  • For \( f(x) = x^2 \), the slope function or derivative \( f'(x) = 2x \) tells us the slope at any point on the curve.
  • So, at \( x = 2 \), the slope is 4, which is the slope of the tangent line there.
  • This slope similarity is seen at \( x = -1 \) with a slope of -2.
Slope is an integral component in understanding graphs and is used across calculus to define and explain the behavior of functions through their tangent lines.

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Most popular questions from this chapter

Let \(f(x)=x^{3}\) and \(P\) be the point \((1,1)\) on the graph of \(f\). (a) Approximate the slope of the line tangent to \(f\) at \(P\) by looking at the slope of the secant line through \(P\) and \(Q\), where \(Q=(1+h, f(1+h)\) ). Calculate the difference quotient for various values of \(h\), both positive and negative. See if your calculator or computer will produce a table of values. (b) Calculate \(f^{\prime}(1)\) by computing the limit of the difference quotient. (c) For what values of \(h\) is the difference quotient greater than \(f^{\prime}(1) ?\) For what values of \(h\) is the difference quotient less than \(f^{\prime}(1) ?\) Make sense out of this by looking at the graph of \(x^{3}\).

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?

An orange is growing on a tree. Assume that the orange is always spherical, and that it has not yet reached its mature size. Its current radius is \(r \mathrm{~cm}\). (a) If the radius increases by \(0.5 \mathrm{~cm}\), what is the corresponding increase in volume? What is \(\frac{\Delta V}{\Delta r}\) ? (b) If the radius of the orange increases by \(\Delta r\), what is the corresponding increase in volume? What is \(\frac{\Delta V}{\Delta r} ?\) (Please simplify your answer.) (c) Show that \(\lim _{\Delta r \rightarrow 0} \frac{\Delta V}{\Delta r}=4 \pi r^{2}\). Conclude that for \(\Delta r\) very small \(\Delta V \approx\left(4 \pi r^{2}\right) \Delta r\). (d) The surface area of a sphere is \(4 \pi r^{2}\). Explain, in terms of an orange, why the approximation \(\Delta V \approx\left(4 \pi r^{2}\right) \Delta r\) make sense.

Using the limit de nition of the derivative, nd \(f^{\prime}(x)\) if \(f(x)=(x-1)^{2}\).

If we have a formula for \(f\) we can get quite good numerical estimates of the slope of the tangent line to \(f\) at a particular point. In this exercise we will do that. Below is the graph of \(y=f(x)=x^{2}-4\) (a) Goal: We want to estimate the value of \(f^{\prime}(2)\), i.e., to approximate the slope of the tangent line to the graph at the point \((2,0)\). To this end, do the following: (b) Goal: We want to estimate the value of \(f^{\prime}(1) ;\) that is, we want to approximate the slope of the tangent line to the graph at the point \((1,-3)\). (c) Goal: We want to estimate the value of \(f^{\prime}(0) ;\) i.e., we want to approximate the slope of the tangent line to the graph at the point \((0,-4)\). i. Sketch the tangent line to the graph at \((0,-4)\). ii. Find the slope of the line through \((0,-4)\) and \((0+h, f(0+h))\). iii. Use your answer to part ii to estimate the slope of the tangent line. (d) Goal: To estimate the value of \(f^{\prime}(\mathrm{c})\) for \(\mathrm{c}\) a constant. i. Find the slope of the line through (c, \(f(\mathrm{c}))\) and \((\mathrm{c}+h, f(\mathrm{c}+h))\). ii. Use your answer to the previous question to nd the slope of the tangent line to \(f\) at \(x=\mathrm{c}\). iii. Sketch the graph of \(f^{\prime}(x)\).
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