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Chapter 3: Problem 61

Finding \(f\) from \(f^{\prime}\) Sketch the graph of \(f^{\prime}(x)=2 .\) Then sketch three possible graphs of \(f\)

Short Answer

Expert verified
Question: Sketch the graph of the derivative function \(f^{\prime}(x) = 2\) and provide three possible graphs of the original function \(f(x)\) considering information from the derivative function. Answer: The graph of the derivative \(f^{\prime}(x)=2\) is a horizontal line at the \(y\)-value 2. Based on the constant positive slope indicated by the derivative, three possible graphs of the original function \(f(x)\) are: 1) \(f(x) = 2x\) (passes through the origin and has a slope of 2). 2) \(f(x) = 2x + c\) (passes through the point \((0, c)\), where \(c\) is an arbitrary constant, and has a slope of 2). 3) \(f(x) = 2x - 2a\) (passes through the point \((a, 0)\), where \(a\) is an arbitrary constant offset from the origin, and has a slope of 2).

Step by step solution

01

Analyze the derivative function \(f^{\prime}(x)=2\)

The derivative function \(f^{\prime}(x)\) is a constant function with the value \(2\). This means that the slope of the tangent lines to the graph of the original function \(f(x)\) will always be \(2\). This information will be useful when sketching the graphs of the original function.
02

Sketch the graph of the derivative \(f^{\prime}(x)=2\)

Since \(f^{\prime}(x)\) is a constant function with the value \(2\), its graph will be a horizontal line at the \(y\)-value 2. The graph will be a straight line that passes through the point \((0, 2)\) and remains parallel to the\(x\)-axis throughout.
03

Sketch possible graphs of the original function \(f(x)\)

To sketch the possible graphs of the original function \(f(x)\), we need to reverse-engineer the derivative information. Since the derivative, and hence the slope of the tangent lines, is always positive, it means that the original function \(f(x)\) is always increasing. Also, the slope of the tangent lines is constant and equal to \(2\), meaning that \(f(x)\) increases linearly. 1) Assume \(f(0)=0\): We can choose one possible graph of \(f(x)\) by assuming it passes through the origin, i.e., \((0, 0)\). With this assumption, the graph of \(f(x)\) will be a straight line passing through the origin with a slope of \(2\). Hence, \(f(x) = 2x\). 2) Assume \(f(0)=c\), where \(c\) is a constant: We can also sketch another graph of \(f(x)\) by assuming it passes through the point \((0, c)\), where \(c\) is an arbitrary constant. The graph will still have a slope of \(2\), and the equation of this line will be \(f(x)=2x+c\). 3) Offset the starting point: We can also sketch a third graph of \(f(x)\) by altering the\(x\)-axis starting point. For example, suppose the graph passes through the point \((a, 0)\), where \(a\) is an arbitrary constant offset from the origin. The equation of this line will still be \(f(x) = 2x - 2a\).

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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 a type of function where every output value is the same regardless of the input. In mathematical terms, if we have a function \(f(x)\), it is a constant function if \(f(x) = c\) where \(c\) is a constant number. This means the graph of a constant function is a horizontal line on the coordinate plane because it doesn't change for different values of \(x\).

In the context of differential equations, derivatives can be constant functions as well. For example, when we have a derivative \(f'(x) = 2\), it tells us that the slope of the tangent at any point of \(f(x)\) is always 2.
  • The graph will be a horizontal line either above or below the \(x\)-axis, depending on the sign of the constant.
  • In our case, we are dealing with a line above the \(x\)-axis because \(f'(x) = 2\) is positive.
This characteristic can significantly simplify the process of sketching graphs, as the behavior of the function is predictable and uniform across its domain.
Tangent Line
A tangent line is a straight line that touches a curve at a single point and has the same slope as the curve at that point. When dealing with functions and their derivatives, the derivative at a specific point gives us the slope of the tangent line.

- For example, if we have a function \(f(x)\) and \(f'(x) = 2\), then at any point on \(f(x)\), the tangent line will have a slope of 2.- This means that if you were standing on the curve, at any point, and wanted to know the angle of the slope beneath you, it would always be 2.

In geometrical visualization, this can be quite helpful for understanding how the function behaves because if the slope \(= 2\) is consistent, then the tangent lines across different points on \(f(x)\) would form parallel lines. This reflects the constant increase or decrease when we analyze the function holistically.
Linear Function
Linear functions are functions that create straight lines when graphed. They have the general form \(f(x) = mx + b\), where \(m\) represents the slope, and \(b\) is the \(y\)-intercept. This means linear functions change at a constant rate, which is designated by the slope \(m\).

Given a derivative like \(f'(x) = 2\), it implies that the original function \(f(x)\) is a linear function because of the constant slope of the tangent. Here the slope \(m = 2\), meaning \(f(x) = 2x + c\), where \(c\) is a constant that could be derived from different initial conditions or given points on the line.
  • If the graph passes through the origin, \(c\) would be zero: \(f(x) = 2x\).
  • If it passes through another point like \((0, c)\), this changes the \(y\)-intercept \(b = c\).
  • Offsetting the line to pass through a point \((a, 0)\) adjusts the equation to \(f(x) = 2x - 2a\).
These options show the flexibility allowed by linear equations when establishing initial conditions or interpreting graphical shifts and transformations.

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Most popular questions from this chapter

Tangent lines and exponentials Assume \(b\) is given with \(b > 0\) and \(b \neq 1 .\) Find the \(y\) -coordinate of the point on the curve \(y=b^{x}\) at which the tangent line passes through the origin. (Source: The College Mathematics Journal, \(28,\) Mar 1997 )

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure that they actually lie on the curve. Confirm your results with a graph. $$x\left(1-y^{2}\right)+y^{3}=0$$

86-89. Second derivatives Find \(\frac{d^{2} y}{d x^{2}}\) for the following functions. $$y=\sin x^{2}$$

Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. A damped oscillator The displacement of a mass on a spring suspended from the ceiling is given by \(y=10 e^{-t / 2} \cos \frac{\pi t}{8}\) a. Graph the displacement function. b. Compute and graph the velocity of the mass, \(v(t)=y^{\prime}(t)\) c. Verify that the velocity is zero when the mass reaches the high and low points of its oscillation.

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$\cos y^{2}+x=e^{y}$$
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