Consider the following statements: 1. \({\rm{y}} = \frac{{{{\rm{e}}^{\rm{x}}} + {{\rm{e}}^{ - {\rm{x}}}}}}{2}\) is an increasing function on [0, ∞). 2. \({\rm{y}} = \frac{{{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}}}}{2}\) is an increasing function on (-∞, ∞). Which of the above statements is/are correct?
Both 1 and 2
Analyzing Increasing Functions Using Derivatives
To determine if a function \(f(x)\) is increasing on a given interval, we typically analyze the sign of its derivative, \(f'(x)\), over that interval.
- If \(f'(x) \ge 0\) for all \(x\) in the interval, then the function \(f(x)\) is increasing on that interval.
- If \(f'(x) > 0\) for all \(x\) in the interval (except possibly at isolated points), then the function \(f(x)\) is strictly increasing on that interval.
Statement 1 Analysis: \({\rm{y}} = \frac{{{{\rm{e}}^{\rm{x}}} + {{\rm{e}}^{ - {\rm{x}}}}}}{2}\) on [0, ∞)
Let the function be \(y_1(x) = \frac{{{\rm{e}}^{\rm{x}}} + {{\rm{e}}^{ - {\rm{x}}}}}}{2}\). This function is also known as the hyperbolic cosine, \(\cosh x\).
We need to find the derivative of \(y_1(x)\) with respect to \(x\): \[ \frac{d}{dx} \left( \frac{{{\rm{e}}^{\rm{x}}} + {{\rm{e}}^{ - {\rm{x}}}}}}{2} \right) \] \[ y_1'(x) = \frac{1}{2} \left( \frac{d}{dx}({{\rm{e}}^{\rm{x}}}) + \frac{d}{dx}({{\rm{e}}^{ - {\rm{x}}}}) \right) \] \[ y_1'(x) = \frac{1}{2} ({{\rm{e}}^{\rm{x}}} + (-{{\rm{e}}^{ - {\rm{x}}}}) \cdot 1) \] \[ y_1'(x) = \frac{{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}}}}{2} \] This derivative is also known as the hyperbolic sine, \(\sinh x\).
Now, we need to check the sign of \(y_1'(x)\) on the interval [0, ∞). For \(x \ge 0\):
- If \(x = 0\), \(y_1'(0) = \frac{{{\rm{e}}^0} - {{\rm{e}}^{-0}}}{2} = \frac{1 - 1}{2} = 0\).
- If \(x > 0\), \({\rm{e}}^{\rm{x}}\) is an increasing function, so for \(x > 0\), \({\rm{e}}^{\rm{x}} > {{\rm{e}}^0} = 1\).
- Also, \({\rm{e}}^{ - {\rm{x}}}\) is a decreasing function, so for \(x > 0\), \({\rm{e}}^{ - {\rm{x}}} < {{\rm{e}}^0} = 1\). More specifically, \(0 < {{\rm{e}}^{ - {\rm{x}}}} < 1\).
Since \({\rm{e}}^{\rm{x}} > 1\) and \(0 < {{\rm{e}}^{ - {\rm{x}}}} < 1\) for \(x > 0\), it follows that \({\rm{e}}^{\rm{x}} > {{\rm{e}}^{ - {\rm{x}}}}\) for \(x > 0\). Therefore, for \(x > 0\), \({\rm{e}}^{\rm{x}} - {{\rm{e}}^{ - {\rm{x}}}} > 0\).
Combining the cases for \(x=0\) and \(x > 0\), we have \(y_1'(x) = \frac{{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}}}}{2} \ge 0\) for all \(x \in [0, \infin;)\). Thus, the function \({\rm{y}} = \frac{{{{\rm{e}}^{\rm{x}}} + {{\rm{e}}^{ - {\rm{x}}}}}}{2}\) is an increasing function on [0, ∞).
Statement 1 is correct.
Statement 2 Analysis: \({\rm{y}} = \frac{{{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}}}}{2}\) on (-∞, ∞)
Let the function be \(y_2(x) = \frac{{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}}}}{2}\). This function is also known as the hyperbolic sine, \(\sinh x\).
We need to find the derivative of \(y_2(x)\) with respect to \(x\): \[ \frac{d}{dx} \left( \frac{{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}}}}{2} \right) \] \[ y_2'(x) = \frac{1}{2} \left( \frac{d}{dx}({{\rm{e}}^{\rm{x}}}) - \frac{d}{dx}({{\rm{e}}^{ - {\rm{x}}}}) \right) \] \[ y_2'(x) = \frac{1}{2} ({{\rm{e}}^{\rm{x}}} - (-{{\rm{e}}^{ - {\rm{x}}}}) \cdot 1) \] \[ y_2'(x) = \frac{{{\rm{e}}^{\rm{x}}} + {{\rm{e}}^{ - {\rm{x}}}}}}{2} \] This derivative is also known as the hyperbolic cosine, \(\cosh x\).
Now, we need to check the sign of \(y_2'(x)\) on the interval (-∞, ∞). For any real number \(x\):
- \({\rm{e}}^{\rm{x}} > 0\)
- \({\rm{e}}^{ - {\rm{x}}} > 0\)
The sum of two positive numbers is always positive. Therefore, \({\rm{e}}^{\rm{x}} + {{\rm{e}}^{ - {\rm{x}}}} > 0\) for all \(x \in (-\infin;, \infin;)\). Thus, \(y_2'(x) = \frac{{{\rm{e}}^{\rm{x}}} + {{\rm{e}}^{ - {\rm{x}}}}}}{2} > 0\) for all \(x \in (-\infin;, \infin;)\). This means the function \({\rm{y}} = \frac{{{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}}}}{2}\) is strictly increasing on (-∞, ∞).
Statement 2 is correct.
Conclusion on Increasing Functions
Based on the analysis of their derivatives:
- Statement 1, concerning \({\rm{y}} = \frac{{{{\rm{e}}^{\rm{x}}} + {{\rm{e}}^{ - {\rm{x}}}}}}{2}\) being increasing on [0, ∞), is correct because its derivative is non-negative on this interval.
- Statement 2, concerning \({\rm{y}} = \frac{{{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}}}}{2}\) being increasing on (-∞, ∞), is correct because its derivative is positive on this interval.
Therefore, both statements are correct.
| Statement | Function \(y(x)\) | Interval | Derivative \(y'(x)\) | Sign on Interval | Increasing? |
|---|---|---|---|---|---|
| 1 | \( \frac{{{\rm{e}}^{\rm{x}}} + {{\rm{e}}^{ - {\rm{x}}}}}}{2} \) | [0, ∞) | \( \frac{{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}}}}{2} \) | \( \ge 0 \) | Yes |
| 2 | \( \frac{{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}}}}{2} \) | (-∞, ∞) | \( \frac{{{\rm{e}}^{\rm{x}}} + {{\rm{e}}^{ - {\rm{x}}}}}}{2} \) | \( > 0 \) | Yes |
Revision Table: Increasing Functions and Derivatives
| Concept | Description | How to Check for Increasing Function |
|---|---|---|
| Increasing Function | A function \(f(x)\) is increasing on an interval if for any \(x_1, x_2\) in the interval, \(x_1 < x_2\) implies \(f(x_1) \le f(x_2)\). | Check if the derivative \(f'(x) \ge 0\) on the interval. |
| Strictly Increasing Function | A function \(f(x)\) is strictly increasing on an interval if for any \(x_1, x_2\) in the interval, \(x_1 < x_2\) implies \(f(x_1) < f(x_2)\). | Check if the derivative \(f'(x) > 0\) on the interval (except possibly at isolated points where \(f'(x)=0\)). |
| Derivative \(f'(x)\) | Represents the instantaneous rate of change of \(f(x)\). Its sign indicates the direction of the function's change. | Calculate using differentiation rules. |
Additional Information: Hyperbolic Functions and Monotonicity
The functions given in the statements are related to hyperbolic functions:
- \( \cosh x = \frac{{{\rm{e}}^{\rm{x}}} + {{\rm{e}}^{ - {\rm{x}}}}}}{2} \)
- \( \sinh x = \frac{{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}}}}{2} \)
The derivative of \( \cosh x \) is \( \sinh x \), and the derivative of \( \sinh x \) is \( \cosh x \).
The graphs of \( \cosh x \) and \( \sinh x \) can also help visualize their increasing nature:
- The graph of \( \cosh x \) is U-shaped, symmetric about the y-axis. Its minimum is at \(x=0\). It is decreasing on (-\infin;, 0] and increasing on [0, ∞).
- The graph of \( \sinh x \) passes through the origin (0,0) and is strictly increasing over its entire domain (-∞, ∞).
This graphical understanding supports our derivative analysis conclusions about the increasing nature of these functions on the specified intervals.
Important Questions from Applications of Derivatives
The function is decreasing on :
The function attains local minimum value at :
What is the maximum value of y?
What is the maximum value of xy ?
\({\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2} - 1}}{{{{\rm{x}}^2} + 1}}\) where x ϵ R
At what value of x does f(x) attain minimum value?
