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Chapter 4: Problem 15

The polynomial \(q(x)=-7 x^{4}+3 x^{2}-x+4\) is an interpolating polynomial for \(g .\) Use \(q\) to approximate (a) \(g(1)[A]\) (b) \(g(2)\) (c) \(g^{\prime}(1)^{[\Omega]}\) (d) \(g^{\prime}(2)\) (e) \(\int_{0}^{1} g(x) d x\), (f) \(\int_{0}^{2} g(x) d x\)

Short Answer

Expert verified
(a) -1, (b) -98, (c) -23, (d) -213, (e) \(\frac{33}{10}\), (f) \(-\frac{134}{5}\).

Step by step solution

01

Evaluate polynomial at x=1 for g(1)[A]

To find \(g(1)[A]\), substitute \(x = 1\) into the polynomial \(q(x) = -7x^4 + 3x^2 - x + 4\). This gives: \[-7(1)^4 + 3(1)^2 - 1 + 4 = -7 + 3 - 1 + 4 = -1\]. Hence, \(g(1)[A] = -1\).
02

Evaluate polynomial at x=2 for g(2)

To find \(g(2)\), substitute \(x = 2\) into the polynomial \(q(x) = -7x^4 + 3x^2 - x + 4\). This gives: \[-7(2)^4 + 3(2)^2 - 2 + 4 = -112 + 12 - 2 + 4 = -98\]. Hence, \(g(2) = -98\).
03

Differentiate polynomial for derivative

First, differentiate \(q(x) = -7x^4 + 3x^2 - x + 4\) to get its derivative. The derivative is: \(q'(x) = -28x^3 + 6x - 1\).
04

Evaluate derivative at x=1 for g'(1)

Substitute \(x = 1\) into the derivative \(q'(x) = -28x^3 + 6x - 1\). This gives: \[-28(1)^3 + 6(1) - 1 = -28 + 6 - 1 = -23\]. Thus, \(g'(1)[\Omega] = -23\).
05

Evaluate derivative at x=2 for g'(2)

Substitute \(x = 2\) into the derivative \(q'(x) = -28x^3 + 6x - 1\). This gives: \[-28(2)^3 + 6(2) - 1 = -224 + 12 - 1 = -213\]. Thus, \(g'(2) = -213\).
06

Integrate polynomial from 0 to 1 for definite integral

To find \(\int_{0}^{1} g(x) \, dx\), we integrate \(q(x)\): \(\int (-7x^4 + 3x^2 - x + 4) \, dx\). This results in: \(-\frac{7}{5}x^5 + x^3 - \frac{1}{2}x^2 + 4x\). Evaluate from 0 to 1: \[-\frac{7}{5}(1)^5 + (1)^3 - \frac{1}{2}(1)^2 + 4(1) = -\frac{7}{5} + 1 - \frac{1}{2} + 4 = \frac{33}{10}\].
07

Integrate polynomial from 0 to 2 for definite integral

To find \(\int_{0}^{2} g(x) \, dx\), evaluate the antiderivative from the previous step at 2 and 0. Compute: \(-\frac{7}{5}(2)^5 + (2)^3 - \frac{1}{2}(2)^2 + 4(2)\) minus the value at 0. This gives: \[-\frac{224}{5} + 8 - 2 + 8 = -\frac{134}{5}\].

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Key Concepts

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

Interpolating Polynomial
An interpolating polynomial is a mathematical function, often used in numerical analysis, that passes through a set of given points or data. The main purpose is to estimate the values of a function at points not specifically given. In our exercise, the polynomial \( q(x) = -7x^4 + 3x^2 - x + 4 \) serves as the interpolating polynomial for the function \( g \). This means it has been specifically designed to fit through certain values derived from \( g \) at exact points, such as those at \( x = 1 \) and \( x = 2 \).

Here鈥檚 how you use an interpolating polynomial:
  • Plug in the desired \( x \) value into the polynomial to get the corresponding \( g(x) \).
  • This helps in approximating the function \( g \) even if it's complex or unknown.
By substituting values into \( q(x) \), you can find approximations for \( g(1) \) and \( g(2) \), offering insights into the behavior of \( g \) without needing its exact form.
Derivative Approximation
Derivative approximation involves finding an estimate for the derivative, or the rate of change, of a function at a certain point. Derivatives are fundamental in understanding how a function behaves, indicating slope and inflection.

In our context, the derivative of the interpolating polynomial \( q(x) = -7x^4 + 3x^2 - x + 4 \) is calculated, resulting in \( q'(x) = -28x^3 + 6x - 1 \). This represents the rate of change of \( q(x) \) at any point \( x \).
  • To approximate \( g'(1) \), substitute \( x = 1 \) into \( q'(x) \) to find \( g'(1) \).
  • Similarly, substitute \( x = 2 \) for \( g'(2) \).
This derivative approximation provides a way to estimate how \( g \) is changing at particular points, adding another layer of comprehension to the behavior of \( g(x) \).
Definite Integral
A definite integral calculates the area under a curve within a given interval. It's a key concept in calculus that represents accumulation, such as distance, area, or total change.

To find the definite integral of the polynomial \( q(x) \) from 0 to 1, and from 0 to 2, you integrate the function and evaluate the limits. The indefinite integral of \( q(x) = -7x^4 + 3x^2 - x + 4 \) is \(-\frac{7}{5}x^5 + x^3 - \frac{1}{2}x^2 + 4x\). Calculating it over the specified intervals gives you:
  • \( \int_{0}^{1} q(x) dx = \frac{33}{10} \)
  • \( \int_{0}^{2} q(x) dx = -\frac{134}{5} \)
This tells us about the net area between the polynomial and the x-axis over the given intervals. Thus, definite integrals help in understanding the cumulative effect over a range of \( x \), providing substantial information about \( g(x) \) and its characteristics.

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Most popular questions from this chapter

Use the composite midpoint rule with 3 subintervals to approximate (a) \(\int_{1}^{3} \ln (\sin (x)) d x\), (b) \(\int_{5}^{7} \sqrt{x \cos x} d x\) (c) \(\int_{1}^{4} \frac{e^{x} \ln (x)}{x} d x^{[A]}\) (d) \(\int_{10}^{13} \sqrt{1+\cos ^{2} x} d x\) (e) \(\int_{\ln 3}^{\ln 7} \frac{e^{x}}{1+x} d x^{[A]}\) (f) \(\int_{0}^{1} \frac{x^{2}-1}{x^{2}+1} d x\)

Derive an approximation formula for the first derivative ber the stencil following these steps. (a) Write down \(L_{1}(x(\theta))=L_{1}\left(x_{0}+\theta h\right)\), the Lagrange form of the interpolating polynomial passing through the points $$ \left(x_{0}, f\left(x_{0}\right)\right) \quad \text { and } \quad\left(x_{0}+h, f\left(x_{0}+h\right)\right) $$ in terms of \(\theta, h,\) and \(x_{0}\). (b) Calculate the derivative \(\frac{d}{d x} L_{1}(x(\theta)) .\) Remember, \(x(\theta)=x_{0}+\theta h,\) and use the chain rule. (c) Substitute \(\theta=\frac{1}{2}\) into your formula from (b) and simplify. [A]

Use adaptive Simpson's method to approximate \(\int_{0}^{1} \ln (x+1) d x\) accurate to within \(10^{-4}\).

The following data give estimates of the integral \(M=\) $$ \begin{array}{l} \int_{0}^{3 \pi / 2} \cos x d x . \\ N(h)=2.356194 \quad N(h / 2)=-0.4879837 \\ N(h / 4)=-0.8815732 \quad N(h / 8)=-0.9709157 \\ \text { Assuming } M-N(h)=K_{1} h^{2}+K_{2} h^{4}+K_{3} h^{6}+\cdots, \text { find } \end{array} $$ a third Richardson's extrapolation for \(M .\) [s]

Use composite Simpson's rule to estimate \(\int_{0}^{1} \ln (x+1) d x\) accurate to within \(10^{-6}\). How many subintervals are needed?
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