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Chapter 4: Problem 30

Suppose that \(f(0)=5\) and that \(f^{\prime}(x)=2\) for all \(x .\) Must \(f(x)=2 x+5\) for all \(x ?\) Give reasons for your answer.

Short Answer

Expert verified
Yes, \( f(x) = 2x + 5 \) for all \( x \) because it satisfies both the derivative condition and the initial condition.

Step by step solution

01

Understand the Function Information

We are given that the value of the function at 0 is 5, i.e., \( f(0) = 5 \). Also, \( f'(x) = 2 \) for all \( x \). This means the derivative of \( f(x) \) is constant.
02

Recall the Derivative Rule

The derivative \( f'(x) = 2 \) suggests that the function \( f(x) \) is of the form \( f(x) = 2x + C \). This is because the derivative of \( 2x \) is 2, and \( C \) is a constant.
03

Use the Initial Condition

Substitute \( x = 0 \) into the expression \( f(x) = 2x + C \) to use the given initial condition. Thus, \( f(0) = 2(0) + C = 5 \), which simplifies to \( C = 5 \).
04

Combine the Information

Now, substitute the value of \( C \) back into the expression for \( f(x) \). Therefore, we get the function \( f(x) = 2x + 5 \).
05

Confirm No Other Functions Fit

The function \( f(x) = 2x + C \) is general for constant derivatives of 2. Since we solved for \( C \) using \( f(0) = 5 \), \( f(x) = 2x + 5 \) is the unique solution given the constraints.

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Key Concepts

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

Derivative
In calculus, the derivative is a fundamental concept that describes the rate of change of a function. Essentially, it provides the slope of the tangent line to the function's graph at any given point. For a function like \( f(x) \), its derivative, denoted as \( f'(x) \), shows how \( f(x) \) changes with respect to \( x \).
If you know the slope of a line, you can understand how steeply the line rises or falls. For instance, if \( f'(x) = 2 \), it means that for each unit increase in \( x \), \( f(x) \) increases by 2 units.
  • The derivative tells us the function's behavior at every point along its curve.
  • A constant derivative, like \( f'(x) = 2 \), indicates a linear, non-changing slope.

Knowing the derivative is vital for understanding the behavior and shape of different functions.
Constant Function
A constant function is a function that assumes the same value no matter the input. In other words, its output is fixed and does not change over its entire domain.
However, in the context of derivatives, a constant derivative like \( f'(x) = 2 \) indicates that the original function, \( f(x) \), is linearly related to \( x \). This means that \( f(x) \) is not a constant function itself, but rather that the rate of change is constant.
  • Constant functions are typically of the form \( f(x) = c \), where \( c \) is a constant value.
  • A component of a linear function has a constant slope if the derivative is constant.
In sum, while a constant function's value itself does not change, having a constant derivative like \( f'(x) = 2 \) points to a line with an unchanging slope.
Initial Condition
Initial conditions are specific values of a function or its derivatives at particular points, typically used to find specific function parameters. In our example, we have \( f(0) = 5 \), which gives us a starting point to determine other constants within the function, such as \( C \) in \( f(x) = 2x + C \).
By substituting \( x = 0 \) into the function, you set the stage for solving for unknown constants:
  • The known value \( f(0) = 5 \) assists in finding the unknown \( C \) when the function \( f(x) \) is expressed in terms of \( x \).
  • Applying initial conditions ensures that we can uniquely determine the specific form of a function, providing precision to our solution.
Ultimately, initial conditions are crucial in obtaining a complete and definite expression for functions in calculus problems.
Linear Function
Linear functions in mathematics are those that produce straight-line graphs. These functions are commonly written in the form \( f(x) = mx + b \), where \( m \) represents the slope and \( b \) is the y-intercept.
In our problem, since the derivative \( f'(x) = 2 \) is constant, \( f(x) \) takes the form \( f(x) = 2x + C \). This is because the derivative of \( mx \) is \( m \), so when \( m = 2 \), each unit increase in \( x \) results in a 2-unit increase in \( f(x) \).
  • Linear functions have constant slopes, reflected by their constant derivative values.
  • The constant \( C \) determines the vertical positioning of the line on a graph.
Therefore, the function \( f(x) = 2x + 5 \) can be considered a linear function that exactly meets the given initial condition \( f(0) = 5 \) and derivative \( f'(x) = 2 \), creating its unique linear equation.

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Most popular questions from this chapter

The 800 -room Mega Motel chain is filled to capacity when the room charge is \(\$ 50\) per night. For each \(\$ 10\) increase in room charge, 40 fewer rooms are filled each night. What charge per room will result in the maximum revenue per night?

a. When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the questions of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity \(v\) can be modeled by the equation $$v=c\left(r_{0}-r\right) r^{2} \mathrm{cm} / \mathrm{sec}, \quad \frac{r_{0}}{2} \leq r \leq r_{0}$$ where \(r_{0}\) is the rest radius of the trachea in centimeters and \(c\) is a positive constant whose value depends in part on the length of the trachea. Show that \(v\) is greatest when \(r=(2 / 3) r_{0} ;\) that is, when the trachea is about \(33 \%\) contracted. The remarkable fact is that X-ray photographs confirm that the trachea contracts about this much during a cough. b. Take \(r_{0}\) to be 0.5 and \(c\) to be \(1,\) and graph \(v\) over the interval \(0 \leq r \leq 0.5 .\) Compare what you see with the claim that \(v\) is at a maximum when \(r=(2 / 3) r_{0}\).

Solve the initial value problems. $$\frac{d y}{d x}=3 x^{-2 / 3}, \quad y(-1)=-5$$

a. Find a curve \(y=f(x)\) with the following properties: i) \(\frac{d^{2} y}{d x^{2}}=6 x\) ii) Its graph passes through the point (0,1) and has a horizontal tangent line there. b. How many curves like this are there? How do you know?

Suppose that $$f(x)=\frac{d}{d x}(1-\sqrt{x}) \text { and } g(x)=\frac{d}{d x}(x+2).$$ Find: a. \(\int f(x) d x\) b. \(\int g(x) d x\) c. \(\int[-f(x)] d x\) d. \(\int[-g(x)] d x\) e. \(\int[f(x)+g(x)] d x\) f. \(\int[f(x)-g(x)] d x\)
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