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Chapter 1: Problem 6

Suppose that \(y=f(x)\) is a differentiable function for which the following information is known: \(f(2)=-3, f^{\prime}(2)=1.5, f^{\prime \prime}(2)=-0.25 .\) a. Is \(f\) increasing or decreasing at \(x=2 ?\) Is \(f\) concave up or concave down at \(x=2 ?\) b. Do you expect \(f(2.1)\) to be greater than \(-3,\) equal to \(-3,\) or less than \(-3 ?\) Why? c. Do you expect \(f^{\prime}(2.1)\) to be greater than 1.5 , equal to 1.5 , or less than 1.5 ? Why? d. Sketch a graph of \(y=f(x)\) near \((2, f(2))\) and include a graph of the tangent line.

Short Answer

Expert verified
At \(x = 2\), \(f\) is increasing and concave down. \(f(2.1)\) is greater than -3. \(f'(2.1)\) is less than 1.5.

Step by step solution

01

Determine if f is increasing or decreasing at x = 2

To determine if the function is increasing or decreasing at a given point, check the sign of the first derivative at that point. Since \(f'(2) = 1.5\) is positive, it means the function is increasing at \(x = 2\).
02

Determine concavity at x = 2

To determine if the function is concave up or concave down at a given point, check the sign of the second derivative at that point. Since \(f''(2) = -0.25\) is negative, it means the function is concave down at \(x = 2\).
03

Estimate f(2.1)

To estimate \(f(2.1)\), use the linear approximation or the tangent line at \(x=2\). The tangent line approximation is \(f(x) \approx f(2) + f'(2)(x - 2)\). Substituting the given values: \[f(2.1) \approx -3 + 1.5(2.1 - 2)\] \[f(2.1) \approx -3 + 1.5(0.1)\] \[f(2.1) \approx -3 + 0.15\] \[f(2.1) \approx -2.85\] Since \( -2.85 > -3\), we expect \(f(2.1)\) to be greater than \(-3\).
04

Estimate f'(2.1)

Since the second derivative \(f''(2) = -0.25\) is negative, it indicates that the slope of the tangent line (i.e., the first derivative) is decreasing around \(x = 2\). Therefore, \(f'(2.1)\) is expected to be less than \(1.5\).
05

Sketch the graph and tangent line

Draw the point (2, -3) on the graph. Since the first derivative is positive, the function is increasing at \(x = 2\). Because the second derivative is negative, the graph is concave down at \(x = 2\). Draw the tangent line at \(x = 2\) with the slope 1.5 passing through the point (2, -3).

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Key Concepts

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

First Derivative
The first derivative of a function, denoted as \(f'(x)\), provides us with the rate at which the function's value is changing at any given point. It essentially tells us the slope of the tangent line to the curve at that point.

In our given problem, we know that \(f'(2) = 1.5\). Since this value is positive, it indicates that the function \(f(x)\) is increasing at \(x = 2\). This means if you were to look at the graph of \(f(x)\) at that point, the curve would be moving upwards.

Understanding whether a function is increasing or decreasing is important for analyzing behaviors and trends in data. If \(f'(x)\) is positive, the function is going up. If \(f'(x)\) is negative, the function is going down.
Second Derivative
The second derivative of a function, denoted as \(f''(x)\), provides information about the concavity of the function. It tells us how the rate of change of the function's slope is behaving.
If \(f''(x) > 0\), the function is concave up, meaning the slope is increasing. If \(f''(x) < 0\), the function is concave down, meaning the slope is decreasing.

In our example, \(f''(2) = -0.25\), which is negative. This indicates that at \(x = 2\), the function \(f(x)\) is concave down. As a result, the shape of the graph at this point would be like an upside-down bowl.

Knowing whether a function is concave up or down helps us understand the nature of its critical points and the intervals of increase or decrease.
Tangent Line Approximation
Tangent line approximation lets us estimate the value of a function near a given point using the tangent line at that point. This is particularly useful when we need to find an approximate value for the function without solving it directly.

Given the information, we have \(f(2) = -3\) and \(f'(2) = 1.5\). The formula for the tangent line is:
\[ f(x) \approx f(a) + f'(a)(x-a) \]

We want to approximate \(f(2.1)\):
\[ f(2.1) \approx f(2) + f'(2)(2.1-2) \approx -3 + 1.5(0.1) \approx -3 + 0.15 \approx -2.85 \]

Since \(-2.85 > -3\), this shows that \(f(2.1)\) is greater than \(f(2)\). This method gives a linear estimation of the function.
Concavity
Concavity describes how the curvature of a function behaves. It is determined by the second derivative, \(f''(x)\). Specifically, it tells us whether the function is curving upwards or downwards:

  • If \(f''(x) > 0\), the function is concave up (shaped like a cup)
  • If \(f''(x) < 0\), the function is concave down (shaped like a cap)


In our example, since \(f''(2) = -0.25\), the function is concave down at \(x = 2\). This means that the function forms a cap shape near this point.

Understanding concavity is critical in predicting the nature of extrema (maxima or minima). It also helps in understanding the inflection points where the function changes concavity.

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Most popular questions from this chapter

The value, \(V,\) of a particular automobile (in dollars) depends on the number of miles, \(m\), the car has been driven, according to the function \(V=h(m)\). a. Suppose that \(h(40000)=15500\) and \(h(55000)=13200\). What is the average rate of change of \(h\) on the interval \([40000,55000],\) and what are the units on this value? b. In addition to the information given in (a), say that \(h(70000)=11100\). Determine the best possible estimate of \(h^{\prime}(55000)\) and write one sentence to explain the meaning of your result, including units on your answer. c. Which value do you expect to be greater: \(h^{\prime}(30000)\) or \(h^{\prime}(80000)\) ? Why? d. Write a sentence to describe the long-term behavior of the function \(V=h(m)\), plus another sentence to describe the long-term behavior of \(h^{\prime}(m) .\) Provide your discussion in practical terms regarding the value of the car and the rate at which that value is changing.

Use linear approximation to approximate \(\sqrt{25.3}\) as follows. Let \(f(x)=\sqrt{x}\). The equation of the tangent line to \(f(x)\) at \(x=25\) can be written in the form \(y=m x+b\). Compute \(m\) and \(b\). \(m=\) \(\square\) \(b=\) \(\square\) Using this find the approximation for \(\sqrt{25.3}\). Answer: \(\square\)

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Let \(g(x)=-\frac{|x+3|}{x+3}\). a. What is the domain of \(g ?\) b. Use a sequence of values near \(a=-3\) to estimate the value of \(\lim _{x \rightarrow-3} g(x),\) if you think the limit exists. If you think the limit doesn't exist, explain why. c. Use algebra to simplify the expression \(\frac{|x+3|}{x+3}\) and hence work to evaluate \(\lim _{x \rightarrow-3} g(x)\) exactly, if it exists, or to explain how your work shows the limit fails to exist. Discuss how your findings compare to your results in (b). (Hint: \(|a|=a\) whenever \(a \geq 0\), but \(|a|=-a\) whenever \(a<0 .)\) d. True or false: \(g(-3)=-1\). Why? e. True or false: \(-\frac{|x+3|}{x+3}=-1 .\) Why? How is this equality connected to your work above with the function \(g\) ? f. Based on all of your work above, construct an accurate, labeled graph of \(y=g(x)\) on the interval \([-4,-2],\) and write a sentence that explains what you now know about \(\lim _{x \rightarrow-3} g(x)\)
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