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Chapter 2: Problem 18

Use the given information to make a good sketch of the function \(f(x)\) near \(x=3\). $$f(3)=3, f^{\prime}(3)=1, \text { inflection point at } x=3, f^{\prime \prime}(x) < 0 \text { for } x > 3$$

Short Answer

Expert verified
Plot the point (3, 3), draw the tangent at x=3 with slope 1, and sketch a curve that transitions from concave up to concave down at the inflection point x=3, curving downward for x > 3.

Step by step solution

01

- Sketch the point at x = 3

Start by plotting the point \(3, 3\) on the graph since \(f(3) = 3\). This is a known point on the function.
02

- Draw the tangent line at x = 3

Since \(f^{\prime}(3) = 1\), the slope of the tangent line at \(x = 3\) is \(1\). The equation of the tangent line is \((y - 3) = 1(x - 3)\), or \(y = x\). Draw this line through the point \(3, 3\).
03

- Identify the inflection point requirements

Since \(x = 3\) is an inflection point, the curve changes concavity at this point. This means the slope of the function transitions from increasing to decreasing or vice versa around \(x = 3\).
04

- Analyze the concavity for x > 3

Given \(f^{\prime\prime}(x) < 0\) for \(x > 3\), the function is concave down for \(x > 3\). Therefore, to the right of \(x = 3\), the graph will curve downwards.
05

- Combine all information for the sketch

Using the plotted point \(3, 3\), the tangent line with slope \(1\), and the concavity information, sketch a smooth curve passing through the point. The curve should transition from concave up to concave down at \(x = 3\) and curve downward for \(x > 3\).

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Key Concepts

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

Inflection Point
An inflection point is where a curve changes its concavity. For our function, there is an inflection point at \(x=3\). At an inflection point, the second derivative \(f''(x)\) either changes from positive to negative or vice versa.

In other words, at \(x=3\), our function transitions.
  • If it was curving upwards (concave up), it starts curving downwards (concave down)
  • If it was curving downwards, it starts curving upwards.
Thus, at \(x = 3\), the concavity of our function changes.
But it's crucial to remember that just having \(f''(3) = 0\) doesn't ensure an inflection point. We must see a change in sign around that point.
Tangent Line
A tangent line touches the function at one exact point, reflecting the slope or rate of change at that point.

For our function near \(x = 3\), the derivative \(f'(3) = 1\) means the slope of the tangent line at \(x = 3\) is 1.

You can use this to form the equation of the tangent line. Here's how:
  • We know it passes through the point \( (3, 3) \).
  • With the slope being 1, we use point-slope form: \((y - y_1) = m(x - x_1) \)

So, \( (y - 3) = 1(x - 3) \), which simplifies to \( y = x \).
Drawing this line through (3, 3) gives a linear representation of our function's immediate behavior at that point.
Concavity
Concavity describes how a function curves. It's determined by the second derivative \(f''(x)\).

For \(x > 3\), we're given \(f''(x) < 0\). This means the function is concave down.

We can see:
  • Concave down: The graph curves like an upside-down cup.\(f''(x) < 0\).
  • Concave up: The graph curves like a right-side-up cup.\(f''(x) > 0\).
For this exercise, starting from \(x = 3\), the function curves downwards. This ensures our sketch beyond \(3\) should show the curve moving downward.

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Most popular questions from this chapter

The graph of each function has one relative extreme point. Find it (giving both \(x\) - and \(y\) -coordinates) and determine if it is a relative maximum or a relative minimum point. Do not include a sketch of the graph of the function. $$g(x)=3+4 x-2 x^{2}$$

Draw the graph of a function \(y=f(x)\) with the stated properties. The function decreases and the slope increases as \(x\) increases. [Note: The slope is negative but becomes less negative.]

The monthly demand equation for an electric utility company is estimated to be $$p=60-\left(10^{-5}\right) x,$$ where \(p\) is measured in dollars and \(x\) is measured in thousands of kilowatt- hours. The utility has fixed costs of 7 million dollars per month and variable costs of \(\$ 30\) per 1000 kilowatt-hours of electricity generated, so the cost function is $$a(x)=7 \cdot 10^{6}+30 x,$$ (a) Find the value of \(x\) and the corresponding price for 1000 kilowatt-hours that maximize the utility's profit. (b) Suppose that rising fuel costs increase the utility's variable costs from \(\$ 30\) to \(\$ 40\), so its new cost function is $$C_{1}(x)=7 \cdot 10^{6}+40 x.$$ Should the utility pass all this increase of \(\$ 10\) per thousand kilowatt- hours on to consumers? Explain your answer.

Each of the graphs of the functions has one relative extreme point. Plot this point and check the concavity there. Using only this information, sketch the graph. [Recall that if \(f(x)=a x^{2}+b x+c,\) then \(f(x)\) has a relative minimum point when \(a>0\) and a relative maximum point when \(a<0.1\) $$f(x)=-x^{2}-8 x-10$$

The graph of each function has one relative extreme point. Find it (giving both \(x\) - and \(y\) -coordinates) and determine if it is a relative maximum or a relative minimum point. Do not include a sketch of the graph of the function. $$f(x)=\frac{1}{4} x^{2}-2 x+7$$
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