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Question

The function f(x) = x2 − 2x is strictly decreasing in the interval

The correct answer is

(−∞, 1)

Finding the Interval of Strictly Decreasing Function

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 \(f(x) = x^2 - 2x\).

Calculating the Derivative

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 \(f'(x) = 2x - 2\).

Determining Critical Points

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.

Testing Intervals for Strict Decrease

The critical point \(x=1\) divides the real number line into two intervals:

  • Interval 1: \( (-\infty, 1) \)
  • Interval 2: \( (1, \infty) \)

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 \( (-\infty, 1) \).

Conclusion

Based on the analysis of the derivative, the function \(f(x) = x^2 - 2x\) is strictly decreasing in the interval \( (-\infty, 1) \).

Revision Table: Function Behavior and Derivative

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.

Additional Information: Applications of Derivatives

Understanding where a function is increasing or decreasing is a fundamental concept in calculus with many applications:

  • Optimization Problems: Finding maximum or minimum values of quantities often involves analyzing where the function changes from increasing to decreasing (local maximum) or decreasing to increasing (local minimum).
  • Graphing Functions: Knowing the intervals of increase and decrease helps sketch the graph of a function accurately, showing its general shape and turning points.
  • Economics: Concepts like marginal cost or marginal revenue, which are derivatives of cost and revenue functions, use the idea of increasing/decreasing rates. For instance, analyzing when marginal profit is decreasing can inform business decisions.
  • Physics: Velocity (the derivative of position) tells us about the rate of change of position. Acceleration (the derivative of velocity) tells us about the rate of change of velocity, and its sign indicates if the velocity is increasing or decreasing.

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.

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Important Questions from Applications of Derivatives

  1. The function is decreasing on :

  2. The function attains local minimum value at :

  3. What is the maximum value of y?

  4. What is the maximum value of xy ?

  5. 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?

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