In which one of the following intervals is the function \(f(x) = \frac{x^3}{3}-\frac{7x^2}{2} + 6x + 5\) decreasing?
(1, 6)
Understanding Decreasing Functions
To determine the intervals where a function is decreasing, we need to analyze the sign of its first derivative. A function \(f(x)\) is decreasing on an interval if its first derivative, \(f'(x)\), is negative (\(f'(x) < 0\)) throughout that interval.
Calculating the First Derivative
The given function is \(f(x) = \frac{x^3}{3}-\frac{7x^2}{2} + 6x + 5\). Let's find the first derivative, \(f'(x)\), using the power rule for differentiation (\(\frac{d}{dx}(x^n) = nx^{n-1}\)) and the rule for constants (\(\frac{d}{dx}(c) = 0\)).
\[f'(x) = \frac{d}{dx}\left(\frac{x^3}{3}-\frac{7x^2}{2} + 6x + 5\right)\]
\[f'(x) = \frac{1}{3} \cdot \frac{d}{dx}(x^3) - \frac{7}{2} \cdot \frac{d}{dx}(x^2) + \frac{d}{dx}(6x) + \frac{d}{dx}(5)\]
\[f'(x) = \frac{1}{3} \cdot (3x^{3-1}) - \frac{7}{2} \cdot (2x^{2-1}) + 6 \cdot (1x^{1-1}) + 0\]
\[f'(x) = \frac{1}{3} \cdot (3x^2) - \frac{7}{2} \cdot (2x) + 6 \cdot (x^0) + 0\]
\[f'(x) = x^2 - 7x + 6\]
So, the first derivative is \(f'(x) = x^2 - 7x + 6\).
Finding Critical Points
Critical points are the points where the first derivative is zero or undefined. In this case, \(f'(x) = x^2 - 7x + 6\) is a polynomial, so it's defined everywhere. We set \(f'(x) = 0\) to find the critical points:
\[x^2 - 7x + 6 = 0\]
We can factor this quadratic equation:
\[(x-1)(x-6) = 0\]
This gives us two critical points:
- \(x - 1 = 0 \implies x = 1\)
- \(x - 6 = 0 \implies x = 6\)
These critical points divide the number line into intervals where the sign of \(f'(x)\) does not change. The intervals are \((-\infty, 1)\), \((1, 6)\), and \((6, \infty)\).
Analyzing Intervals for Decreasing Behavior
Now, we test the sign of \(f'(x) = x^2 - 7x + 6\) in each of these intervals to see where it is negative.
| Interval | Test Value (\(x\)) | \(f'(x) = x^2 - 7x + 6\) | Sign of \(f'(x)\) | Behavior of \(f(x)\) |
|---|---|---|---|---|
| \((-\infty, 1)\) | 0 | \(0^2 - 7(0) + 6 = 6\) | Positive (\(> 0\)) | Increasing |
| \((1, 6)\) | 2 | \(2^2 - 7(2) + 6 = 4 - 14 + 6 = -4\) | Negative (\(< 0\)) | Decreasing |
| \((6, \infty)\) | 7 | \(7^2 - 7(7) + 6 = 49 - 49 + 6 = 6\) | Positive (\(> 0\)) | Increasing |
Based on the analysis, the function \(f(x)\) is decreasing in the interval where \(f'(x) < 0\), which is \((1, 6)\).
Revision Table: Analyzing Function Behavior
| Condition on \(f'(x)\) | Behavior of \(f(x)\) |
|---|---|
| \(f'(x) > 0\) | Increasing |
| \(f'(x) < 0\) | Decreasing |
| \(f'(x) = 0\) | Critical Point (Potential local max/min or inflection point) |
Additional Information on Function Analysis
Understanding intervals where a function is decreasing or increasing is a fundamental part of curve sketching and analyzing function behavior. Here are some related concepts:
- Increasing Function: A function is increasing on an interval if its first derivative is positive (\(f'(x) > 0\)) on that interval.
- Local Extrema: Local maximum or minimum values can occur at critical points (where \(f'(x) = 0\) or is undefined).
- First Derivative Test: This test uses the sign change of \(f'(x)\) around a critical point to determine if it's a local maximum, local minimum, or neither. If \(f'(x)\) changes from positive to negative, it's a local maximum. If it changes from negative to positive, it's a local minimum.
- Second Derivative Test: This test uses the sign of the second derivative \(f''(x)\) at a critical point \(c\) (where \(f'(c)=0\)) to determine local extrema. If \(f''(c) > 0\), there's a local minimum. If \(f''(c) < 0\), there's a local maximum.
- Concavity: The second derivative \(f''(x)\) tells us about the concavity of the function's graph. If \(f''(x) > 0\), the function is concave up. If \(f''(x) < 0\), it's concave down.
Similar Questions
What is the slope of the tangent of y = cos -1 (cos x) at x = \(-\frac{\pi}{5}\) ?
The maximum value of \(\sin \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right) + \cos \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right)\) in the interval \(\left( {0,\frac{{\rm{\pi }}}{2}} \right)\) is attained at
What is the minimum value of [x(x – 1) + 1] 1/3 , where 0 ≤ x ≤ 1?
Let \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}} + \frac{1}{{\rm{x}}}\) , where x ∈ (0, 1). Then which one of the following is correct?
What is the least value of m(θ) ?
Let P be the median, Q be the mean and R be the mode of observations x1, x2, x3, .....xn. Let \(S=\sum_{i=1}^n\left(2 x_i-a\right)^2\) S takes minimum value, when a is equal to
What is the maximum value of xy ?
Consider the following statements in respect of the function f(x) = sin x:
1. f(x) increases in the interval (0, π).
2. f(x) decreases in the interval \(\left(\dfrac{5\pi}{2},3\pi\right).\)
Which of the above statements is/are correct?
What is the maximum area of a triangle that can be inscribed in a circle of radius a?
What is the maximum value of sin 2x ⋅ cos 2x?
Important Questions from Applications of Derivatives
What is the slope of the tangent of y = cos -1 (cos x) at x = \(-\frac{\pi}{5}\) ?
The maximum value of \(\sin \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right) + \cos \left( {{\rm{x}} + \frac{{\rm{\pi }}}{6}} \right)\) in the interval \(\left( {0,\frac{{\rm{\pi }}}{2}} \right)\) is attained at
The derivative of the function y = 3|x| + 1 at the point x = 0 is
Given that f(x) = x 1/x , x > 0 has the maximum value at x = e, then
What is the minimum value of [x(x – 1) + 1] 1/3 , where 0 ≤ x ≤ 1?
