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The interval of increases of the function \(f(x)=x-e^{x}\) \(+\tan \left(\frac{2 \pi}{7}\right)\) is equal to (a) \((0, \infty)\) (b) \((-\infty, 0)\) (c) \((1, \infty)\) (d) \((-\infty,-1)\)

Short Answer

Expert verified
The correct answer option is (b).

Step by step solution

01

Differentiate The Function

First, let's compute the derivative of the function \(f(x) = x - e^x + \tan\left(\frac{2\pi}{7}\right)\). Note that the derivative of \(x\) is 1, the derivative of \(e^x\) is \(e^x\), and since \(\tan\left(\frac{2\pi}{7}\right)\) is a constant number, its derivative is 0. Therefore, the derivative of \(f(x)\), denoted as \(f'(x)\), is \(f'(x) = 1 - e^x + 0 = 1 - e^x\).
02

Determine Intervals Where The Derivative Is Greater Than Zero

Next, we want to find the interval where the derivative \(f'(x)\) is greater than 0, i.e., when \(1 - e^x > 0\). Solving for \(x\), we get \(e^x < 1\). Now we apply the logarithm to both sides (since the logarithm is an increasing function, inequality direction will not change), we get \(x < \ln(1)\). We know that \(\ln(1) = 0\), so the inequality simplifies to \(x < 0\).
03

Compare The Solution With The Answer Options

Lastly, we see that the solution \(x < 0\) corresponds to the open interval \((-∞, 0)\). Inspecting the options given in the problem, we see that this is consistent with option (b).

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Key Concepts

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

Derivative
The derivative is a cornerstone of Differential Calculus. It measures how a function changes as its input changes—essentially describing the rate of change or the function's slope at a given point. For a function like \( f(x) = x - e^{x} + \tan \left(\frac{2\pi}{7}\right) \), the derivative \( f'(x) \) can tell us about how the function behaves across different values of \( x \).
To find \( f'(x) \), we differentiate each part:
  • Derivative of \( x \) is 1 since it increases by 1 for each unit increase in \( x \).
  • Derivative of \( e^{x} \) is \( e^{x} \). This exponential component grows rapidly, influencing the overall behavior significantly.
  • Since \( \tan\left(\frac{2\pi}{7}\right) \) is a constant, its derivative is 0.
By combining these, we get \( f'(x) = 1 - e^{x} \). This derivative helps us understand whether the function is increasing or decreasing, depending on the sign of \( f'(x) \).
Interval of Increase
An interval of increase for a function refers to the range of \( x \) values where the function is rising. If \( f'(x) > 0 \) over an interval, then \( f(x) \) increases within that interval. In our example, determining these intervals is crucial to understanding the behavior of the function \( f(x) = x - e^{x} + \tan\left(\frac{2\pi}{7}\right) \).
We found that the derivative \( f'(x) = 1 - e^{x} \) should be greater than zero to signify an interval of increase. This results in the inequality \( 1 - e^{x} > 0 \), which simplifies further to \( e^{x} < 1 \).
By taking the natural logarithm of both sides, we have \( x < \ln(1) \), meaning \( x < 0 \), since \( \ln(1) = 0 \). Therefore, the function increases in the interval \((-\infty, 0)\). Recognizing this interval helps identify where the function is rising in value.
Exponential Functions
Exponential functions feature prominently in calculus and have a unique property: the rate of growth is proportional to the function's current value. The function \( e^{x} \) is a classic example, frequently showing in calculus due to its distinctive property where the derivative is equal to the function itself, \( e^{x} \).
This exponential growth plays a significant role in our example, \( f(x) = x - e^{x} + \tan\left(\frac{2\pi}{7}\right) \), affecting the function's behavior more rapidly as \( x \) increases.
Understanding \( e^{x} \) is essential. Unlike polynomial functions, exponential functions grow without bound at a faster rate, making them more impactful on derivatives and resulting mathematical behavior.
In practical terms, evaluating \( e^{x} \) for positive \( x \) shows rapid increase, while \( e^{x} \) for negative \( x \) approaches zero, hence impacting the signs of expressions like \( 1 - e^{x} \) significantly and altering where \( f(x) \) increases.

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