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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, because \( f(x) = 2x + 5 \) satisfies both \( f'(x) = 2 \) and \( f(0) = 5 \).

Step by step solution

01

Understanding the Problem

We are given that \( f(0) = 5 \) and \( f'(x) = 2 \) for all \( x \). The task is to determine if \( f(x) \) is equal to \( 2x + 5 \) for all \( x \).
02

Analyze the Derivative of f(x)

Since \( f'(x) = 2 \) is constant for all \( x \), it means that the function \( f(x) \) is linear. Therefore, \( f(x) = 2x + C \), where \( C \) is a constant to be determined.
03

Use the Given Condition f(0) = 5

We know \( f(0) = 5 \). Substitute \( x = 0 \) into the equation \( f(x) = 2x + C \): \[ f(0) = 2(0) + C = C = 5 \]. Thus, \( C = 5 \).
04

Write the Final Function

Substitute \( C = 5 \) back into \( f(x) = 2x + C \). The function becomes \( f(x) = 2x + 5 \).
05

Verify the Conclusion

The derived function \( f(x) = 2x + 5 \) satisfies both \( f'(x) = 2 \) for all \( x \) and \( f(0) = 5 \). Thus, this is the correct function.

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Key Concepts

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

Constant function
A constant function is one of the simplest types of functions in mathematics. It is a function where the output value remains the same regardless of the input. In mathematical terms, a constant function is expressed as \( f(x) = c \) where \( c \) is a constant.
This means that no matter what value of \( x \) you plug into a constant function, the output will always be \( c \). It's like having a straight horizontal line on a graph.
  • The derivative of a constant function is always zero, which means its rate of change is zero.
  • Constant functions are always parallel to the x-axis and never intersect it.
In the case of integration, the integral of a constant function \( c \) with respect to \( x \) is \( cx + C \), where \( C \) represents the constant of integration.
Linear function
Linear functions are a step up in complexity from constant functions. They have the form \( f(x) = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.
This function represents a straight line on a graph. Unlike a constant function's horizontal line, a linear function can have a slope that makes the line rise or fall.
  • The slope \( m \) indicates the rate of change; a positive \( m \) means the line ascends, while negative \( m \) means it descends.
  • The y-intercept \( b \) indicates where the line crosses the y-axis.
The derivative of a linear function is the constant \( m \), reflecting the constant rate of change. This helps us determine that if \( f'(x) = 2 \), then the function is linear with a slope of 2.
Integration
Integration is the mathematical process of finding the integral of a function. It is essentially the reverse operation of differentiation.
When you integrate a function, you are looking for the function with a derivative that matches the original function. For example, the integral of \( 2 \) with respect to \( x \) is \( 2x + C \), where \( C \) is the integration constant.
  • Integration can be thought of as finding the area under the curve of a graphed function.
  • Definite integration results in a numerical area value, while indefinite integration includes the \( C \).
In our exercise, integrating \( f'(x) = 2 \) gives us \( f(x) = 2x + C \), which aligns with finding a function whose derivative is a constant.
Initial condition
Initial conditions are critical in determining the specific form of a function when solving differential equations.
These conditions provide a specific value for the function at a particular point, allowing us to find the constant of integration. In our exercise, \( f(0) = 5 \) serves as an initial condition, telling us how the line \( f(x) = 2x + C \) should be positioned on the coordinate plane.
  • Substituting the initial condition into the integrated function helps solve for \( C \), ensuring the function matches specified conditions.
  • Without an initial condition, there is an infinite number of functions that could satisfy a given differential equation.
Thus, by applying the initial condition \( f(0) = 5 \), we determined that \( C = 5 \), leading to the complete function \( f(x) = 2x + 5 \). This process ensures the solution is unique and satisfies all given parameters.

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Most popular questions from this chapter

You will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Perform the following steps. a. Plot the function over the interval to see its general behavior there. b. Find the interior points where \(f^{\prime}=0 .\) (In some exercises, you may have to use the numerical equation solver to approximate a solution.) You may want to plot \(f^{\prime}\) as well. c. Find the interior points where \(f^{\prime}\) does not exist. d. Evaluate the function at all points found in parts (b) and (c) and at the endpoints of the interval. e. Find the function's absolute extreme values on the interval and identify where they occur. $$f(x)=2+2 x-3 x^{2 / 3}, \quad[-1,10 / 3]$$

You will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Perform the following steps. a. Plot the function over the interval to see its general behavior there. b. Find the interior points where \(f^{\prime}=0 .\) (In some exercises, you may have to use the numerical equation solver to approximate a solution.) You may want to plot \(f^{\prime}\) as well. c. Find the interior points where \(f^{\prime}\) does not exist. d. Evaluate the function at all points found in parts (b) and (c) and at the endpoints of the interval. e. Find the function's absolute extreme values on the interval and identify where they occur. $$f(x)=-x^{4}+4 x^{3}-4 x+1, \quad[-3 / 4,3]$$

Find the inflection points (if any) on the graph of the function and the coordinates of the points on the graph where the function has a local maximum or local minimum value. Then graph the function in a region large enough to show all these points simultaneously. Add to your picture the graphs of the function's first and second derivatives. How are the values at which these graphs intersect the \(x\)-axis related to the graph of the function? In what other ways are the graphs of the derivatives related to the graph of the function? $$y=x^{5}-5 x^{4}-240$$

You will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Perform the following steps. a. Plot the function over the interval to see its general behavior there. b. Find the interior points where \(f^{\prime}=0 .\) (In some exercises, you may have to use the numerical equation solver to approximate a solution.) You may want to plot \(f^{\prime}\) as well. c. Find the interior points where \(f^{\prime}\) does not exist. d. Evaluate the function at all points found in parts (b) and (c) and at the endpoints of the interval. e. Find the function's absolute extreme values on the interval and identify where they occur. $$f(x)=\pi x^{2} e^{-3 x / 2}, \quad[0,5]$$

Solve the initial value problems in Exercises. $$\frac{d^{2} y}{d x^{2}}=2-6 x ; \quad y^{\prime}(0)=4, \quad y(0)=1$$
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