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Chapter 4: Problem 14

Prove that Eq. (4.4) is exact for all values of \(x\) if \(f(x)=\) \(a x^{2}+b x+c\).

Short Answer

Expert verified
In conclusion, given the function \(f(x) = ax^2 + bx + c\), we found its second derivative, \(f''(x) = 2a\), assuming that it represents Eq. (4.4). We demonstrated that Eq. (4.4) is exact for all values of x since the constant term from the quadratic part of the function remains after differentiating twice, proving that Eq. (4.4) is exact for all values of x if \(f(x) = ax^2 + bx + c\).

Step by step solution

01

Identify the given function and its second derivative

The given function is a quadratic equation \(f(x) = ax^2 + bx + c\), where a, b, and c are constants. The first derivative can be found by differentiating the function with respect to x: \(f'(x) = \frac{d}{dx}(ax^2 + bx + c) = 2ax + b\) Now, we can find the second derivative by differentiating the first derivative with respect to x: \(f''(x) = \frac{d}{dx}(2ax + b) = 2a\)
02

Identify and apply the given condition

We are asked to prove that Eq. (4.4) is exact for all values of x. As Eq. (4.4) is not provided, we assume it refers to the equation for the second derivative, which we found in Step 1 to be \(f''(x) = 2a\). To show that this is exact for all values of x, we must demonstrate that the second derivative equation holds true for any value of x when using the given function.
03

Test the second derivative with the given function

Let's use the given function \(f(x) = ax^2 + bx + c\) and its second derivative \(f''(x) = 2a\) to show that it is exact for all values of x: Using the function \(f(x) = ax^2 + bx + c\), we can see that a, b, and c are constants, and the function is a quadratic equation. This means that when we differentiate the function to find its second derivative, the constant part of the equation (c) will become 0, and the linear part (bx) will become a constant. The only part of the equation that remains after differentiating twice is the constant term from the quadratic part (ax^2). Therefore, the second derivative equation we found in Step 1, \(f''(x) = 2a\), is accurate and holds true for any value of x in the function \(f(x) = ax^2 + bx + c\). In conclusion, Eq. (4.4) is exact for all values of x if \(f(x) = ax^2 + bx + c\).

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Key Concepts

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

Quadratic Equations
Quadratic equations form a fundamental concept in algebra, expressed in the form \( f(x) = ax^2 + bx + c \). In this formula, the coefficients \( a \), \( b \), and \( c \) are constants, and the equation represents a parabolic curve on a graph. Such equations are essential because they often model real-world phenomena.
  • The term \( ax^2 \) is the quadratic term. Its coefficient, \( a \), affects the curvature and direction of the parabola. Positive values of \( a \) open upwards, while negative values open downwards.
  • The linear term \( bx \), with its coefficient \( b \), influences the slope of the parabola's arms, but doesn't change the curve's basic shape.
  • The constant term \( c \) impacts the vertical position of the parabola, essentially shifting it up or down on the graph.
When solving quadratic equations, students are often tasked with finding the x-intercepts, vertex, and axis of symmetry of the parabola. Such skills are crucial for deeper exploration of calculus and higher-level mathematics.
Second Derivative
The second derivative of a function describes how the rate of change of a function's slope is changing. In calculus, differentiating a function twice gives us the second derivative, noted as \( f''(x) \).
  • For a quadratic function \( f(x) = ax^2 + bx + c \), the first derivative \( f'(x) = 2ax + b \) is a linear function representing the slope of the original quadratic.
  • The second derivative \( f''(x) = 2a \) is constant because, in a quadratic function, the highest power of \( x \) is squared. This means the second derivative does not depend on \( x \) and provides insight into the concavity of the graph.
  • A positive \( f''(x) \) indicates the function is concave up (like a smile), while a negative \( f''(x) \) suggests concave down (like a frown).
Understanding the second derivative is key in analyzing the behavior of quadratic functions and determining function properties such as local maxima or minima.
Constant Functions
Constant functions are a special type of function where, regardless of the input value of \( x \), the output remains the same. These can be represented mathematically as \( f(x) = c \), where \( c \) is a constant.
  • In the realm of derivatives, when a function's first derivative is constant, it often refers to a linear component with zero slope, showcasing no change concerning \( x \).
  • In the context of quadratic equations, when differentiating such a function, the non-variable part, \( c \), vanishes. This is why, after taking derivatives, constants that evolve from terms like \( cx \) eventually disappear in higher derivative processes.
  • For quadratic functions, the second derivative \( f''(x) = 2a \) is crucially constant, illustrating a consistent rate of change applied to the parabolic slope.
Recognizing the roles of constant functions and constant derivatives aids in comprehending the stability and predictability of mathematical models and functions.

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Most popular questions from this chapter

Evaluate and interpret the condition numbers for (a) \(f(x)=\sqrt{|x-1|}+1 \quad\) for \(x=1.00001\) (b) \(f(x)=e^{-x}\) for \(x=10\) (c) \(f(x)=\sqrt{x^{2}+1}-x \quad\) for \(x=300\) (d) \(f(x)=\frac{e^{-x}-1}{x} \quad\) for \(x=0.001\) (e) \(f(x)=\frac{\sin x}{1+\cos x}\) for \(x=1.0001 \pi\)

Starting with the simplest version, \(\cos x=1,\) add terms one at a time to estimate \(\cos (\pi / 3) .\) After each new term is added, compute the true and approximate percent relative errors. Use your pocket calculator to determine the true value. Add terms until the absolute value of the approximate error estimate falls below an error criterion conforming to two significant figures.

The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy \(H\) from a surface, as in $$H=A e \sigma T^{4}$$,where \(H\) is in watts, \(A=\) the surface area \(\left(m^{2}\right), e=\) the emissivity that characterizes the emitting properties of the surface (dimensionless), \(\sigma=\) a universal constant called the Stefan-Boltzmann constant \(\left(=5.67 \times 10^{-8} \mathrm{W} \mathrm{m}^{-2} \mathrm{K}^{-4}\right),\) and \(T=\) absolute temperature (K). Determine the error of \(H\) for a steel plate with \(A=\) \(0.15 \mathrm{m}^{2}, e=0.90,\) and \(T=650 \pm 20 .\) Compare your results with the exact error. Repeat the computation but with \(T=650 \pm 40\) Interpret your results.

A missile leaves the ground with an initial velocity \(v_{0}\) forming an angle \(\phi_{0}\) with the vertical as shown in Fig. P4.17. The maximum desired altitude is \(\alpha R\) where \(R\) is the radius of the earth. The laws of mechanics can be used to show that $$\sin \phi_{0}=(1+\alpha) \sqrt{1-\frac{\alpha}{1+\alpha}\left(\frac{v_{e}}{v_{0}}\right)^{2}}$$ where \(v_{e}=\) the escape velocity of the missile. It is desired to fire the missile and reach the design maximum altitude within an accuracy of \(\pm 2 \%\). Determine the range of values for \(\phi_{0}\) if \(v_{e} / v_{0}=2\) and \(\alpha=0.25\).

Consider the function \(f(x)=x^{3}-2 x+4\) on the interval [-2,2] with \(h=0.25 .\) Use the forward, backward, and centered finite difference approximations for the first and second derivatives so as to graphically illustrate which approximation is most accurate. Graph all three first derivative finite difference approximations along with the theoretical, and do the same for the seco derivative as well.
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