Problem 21 (a) Evaluate the function \(f(x)... [FREE SOLUTION]

Chapter 1: Problem 21

(a) Evaluate the function \(f(x)=x^{2}-\left(2^{x} / 1000\right)\) for \(x=1,0.8,0.6,0.4,0.2,0.1,\) and \(0.05,\) and guess the value of \(\lim _{x \rightarrow 0}\left(x^{2}-\frac{2^{x}}{1000}\right)\) (b) Evaluate \(f(x)\) for \(x=0.04,0.02,0.01,0.005,0.003\) and \(0.001 .\) Guess again.

Short Answer

Expert verified

\lim_{x \rightarrow 0}(x^{2} - \frac{2^{x}}{1000}) \approx -0.001.

Step by step solution

Evaluate the Function for Given Values of x in Part (a)

We need to evaluate the function \( f(x) = x^2 - \frac{2^x}{1000} \) for the values \( x = 1, 0.8, 0.6, 0.4, 0.2, 0.1, \) and \( 0.05 \). - For \( x = 1 \), \[ f(1) = 1^2 - \frac{2^1}{1000} = 1 - 0.002 = 0.998. \]- For \( x = 0.8 \), \[ f(0.8) = 0.8^2 - \frac{2^{0.8}}{1000} \approx 0.64 - 0.001734 \approx 0.638266. \]- For \( x = 0.6 \), \[ f(0.6) = 0.6^2 - \frac{2^{0.6}}{1000} \approx 0.36 - 0.001515 \approx 0.358485. \]- For \( x = 0.4 \), \[ f(0.4) = 0.4^2 - \frac{2^{0.4}}{1000} \approx 0.16 - 0.001319 \approx 0.158681. \]- For \( x = 0.2 \), \[ f(0.2) = 0.2^2 - \frac{2^{0.2}}{1000} \approx 0.04 - 0.001148 \approx 0.038852. \]- For \( x = 0.1 \), \[ f(0.1) = 0.1^2 - \frac{2^{0.1}}{1000} \approx 0.01 - 0.001072 \approx 0.008928. \]- For \( x = 0.05 \), \[ f(0.05) = 0.05^2 - \frac{2^{0.05}}{1000} \approx 0.0025 - 0.001035 \approx 0.001465. \]

Guess the Limit from Part (a) Observations

Based on the computed values of the function as \( x \) approaches zero, we observe that \( f(x) \) approaches a value closer to zero. Therefore, we can guess that: \[ \lim_{x \rightarrow 0}\left(x^{2} - \frac{2^{x}}{1000}\right) \approx 0. \]

Evaluate the Function for Additional Values of x in Part (b)

Now evaluate \( f(x) \) for \( x = 0.04, 0.02, 0.01, 0.005, 0.003, \) and \( 0.001 \).- For \( x = 0.04 \), \[ f(0.04) = 0.04^2 - \frac{2^{0.04}}{1000} \approx 0.0016 - 0.001028 \approx 0.000572. \]- For \( x = 0.02 \), \[ f(0.02) = 0.02^2 - \frac{2^{0.02}}{1000} \approx 0.0004 - 0.001014 \approx -0.000614. \]- For \( x = 0.01 \), \[ f(0.01) = 0.01^2 - \frac{2^{0.01}}{1000} \approx 0.0001 - 0.001007 \approx -0.000907. \]- For \( x = 0.005 \), \[ f(0.005) = 0.005^2 - \frac{2^{0.005}}{1000} \approx 0.000025 - 0.001003 \approx -0.000978. \]- For \( x = 0.003 \), \[ f(0.003) = 0.003^2 - \frac{2^{0.003}}{1000} \approx 0.000009 - 0.001002 \approx -0.000993. \]- For \( x = 0.001 \), \[ f(0.001) = 0.001^2 - \frac{2^{0.001}}{1000} \approx 0.000001 - 0.0010007 \approx -0.0009997. \]

Guess the Limit from Part (b) Observations

The function \( f(x) \) becomes negative very quickly as \( x \) gets close to zero and approaches a value around \(-0.001\). Therefore, combining both parts, we can refine our guess:\[ \lim_{x \rightarrow 0}\left(x^{2} - \frac{2^{x}}{1000}\right) \approx -0.001. \]

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Key Concepts

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

Function Evaluation

In calculus, evaluating a function involves finding the output of a function for certain inputs. This exercise requires evaluating the function \( f(x) = x^2 - \frac{2^x}{1000} \) at various values of \( x \). When dealing with function evaluation:

Start by substituting each given input value into the function.
Simplify the equation as needed, using exponentiation for \( 2^x \) and arithmetic operations to get the final result.

Understanding how to substitute and simplify allows you to determine how the function behaves at specific values. For example, with \( x = 1 \), you simply calculate \( 1^2 - \frac{2^1}{1000} = 0.998 \). This straightforward process helps interpret the function's behavior without more complex calculations.

Limit Calculation

Limit calculation is a fundamental concept in calculus representing the value that a function approaches as the input approaches a particular point. The exercise focuses on evaluating the limit as \( x \) approaches 0 for the function \( x^2 - \frac{2^x}{1000} \).Understanding limits involves:

Observing the trends of the function outputs as \( x \) values get closer to the point of interest.
Assessing whether the function's output consistently approaches a specific number.

In this exercise, initial values suggest the function approaches 0, implying that as \( x \to 0 \), the limit might be 0. However, upon further evaluation, the additional points show that the function becomes negative as \( x \) becomes extremely small, and the limit refines to close to -0.001.

Approximation Techniques

Approximation techniques are often used when the exact solution is inconvenient or impossible to determine. In context, looking at sequences of function values as \( x \) approaches 0 is a form of numerical approximation of the limit.Applying approximation techniques works by:

Using a sequence of approximations for \( x \) that converge on the limit point.
Observing the changes in the function value to estimate the limiting behavior.

The approximation process involves checking function values for decreasing \( x \), such as 0.04, 0.02, and so forth, to discern the trend toward the limit. This technique helps solidify the guessed limit, counteracting the potential for large errors in singular analytical evaluations.

91影视

(a) Evaluate the function \(f(x)=x^{2}-\left(2^{x} / 1000\right)\) for \(x=1,0.8,0.6,0.4,0.2,0.1,\) and \(0.05,\) and guess the value of \(\lim _{x \rightarrow 0}\left(x^{2}-\frac{2^{x}}{1000}\right)\) (b) Evaluate \(f(x)\) for \(x=0.04,0.02,0.01,0.005,0.003\) and \(0.001 .\) Guess again.

Short Answer

Step by step solution

Evaluate the Function for Given Values of x in Part (a)

Guess the Limit from Part (a) Observations

Evaluate the Function for Additional Values of x in Part (b)

Guess the Limit from Part (b) Observations

Key Concepts

Function Evaluation

Limit Calculation

Approximation Techniques

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