The function f(x) = x2 − 2x is strictly decreasing in the interval
(−∞, 1)
To determine the interval where a function is strictly decreasing, we need to analyze its derivative. A function \(f(x)\) is strictly decreasing on an interval if its derivative \(f'(x)\) is negative on that interval.
The given function is \(f(x) = x^2 - 2x\).
First, let's find the derivative of \(f(x)\) with respect to \(x\). We use the power rule for differentiation, which states that \(\frac{d}{dx}(x^n) = nx^{n-1}\).
For \(f(x) = x^2 - 2x\):
\(f'(x) = \frac{d}{dx}(x^2 - 2x)\)
\(f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(2x)\)
\(f'(x) = 2x^{2-1} - 2x^{1-1}\)
\(f'(x) = 2x^1 - 2x^0\)
\(f'(x) = 2x - 2(1)\)
\(f'(x) = 2x - 2\)
So, the derivative of the function is \(f'(x) = 2x - 2\).
Critical points are where the derivative is zero or undefined. These points can indicate where the function might change from increasing to decreasing or vice versa. To find the critical points, we set \(f'(x) = 0\).
\(2x - 2 = 0\)
\(2x = 2\)
\(x = 1\)
The critical point is \(x = 1\). This point divides the number line into intervals. We will examine the sign of \(f'(x)\) in each interval.
The critical point \(x=1\) divides the real number line into two intervals:
We need to choose a test value within each interval and evaluate the sign of \(f'(x)\) at that point.
| Interval | Test Value (\(x\)) | \(f'(x) = 2x - 2\) | Sign of \(f'(x)\) | Behavior of \(f(x)\) |
|---|---|---|---|---|
| \( (-\infty, 1) \) | 0 | \(2(0) - 2 = -2\) | Negative (\( < 0 \)) | Strictly Decreasing |
| \( (1, \infty) \) | 2 | \(2(2) - 2 = 4 - 2 = 2\) | Positive (\( > 0 \)) | Strictly Increasing |
From the table, we see that \(f'(x)\) is negative in the interval \( (-\infty, 1) \). Therefore, the function \(f(x) = x^2 - 2x\) is strictly decreasing in the interval \( (-\infty, 1) \).
Based on the analysis of the derivative, the function \(f(x) = x^2 - 2x\) is strictly decreasing in the interval \( (-\infty, 1) \).
| Condition on \(f'(x)\) | Behavior of \(f(x)\) | Description |
|---|---|---|
| \(f'(x) > 0\) | Strictly Increasing | The function values increase as \(x\) increases. |
| \(f'(x) < 0\) | Strictly Decreasing | The function values decrease as \(x\) increases. |
| \(f'(x) = 0\) | Stationary Point (Potential Max/Min) | The function's slope is momentarily zero. |
| \(f'(x) \ge 0\) | Increasing | The function values do not decrease as \(x\) increases. |
| \(f'(x) \le 0\) | Decreasing | The function values do not increase as \(x\) increases. |
Understanding where a function is increasing or decreasing is a fundamental concept in calculus with many applications:
The first derivative test is a key tool for finding local extrema (maxima and minima) by examining the sign change of the derivative around critical points. If \(f'(x)\) changes from positive to negative at a critical point \(c\), \(f(c)\) is a local maximum. If \(f'(x)\) changes from negative to positive at \(c\), \(f(c)\) is a local minimum.
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 ?
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?