Increasing Intervals are the ranges of values in which the function INCREASES with respect to the increase in \(x\). Conversely, Decreasing Intervals are the ranges of values in which the function DECREASES with respect to the increase in \(x\).
In order to determine whether a function is increasing or decreasing, we can first find its derivative. We can then conduct the First Derivative Test. This involves solving \(f'(x) = 0\) or \(f'(x) = \text{Undefined}\) by finding test \(x\)-values which can indicate the intervals the function is increasing or decreasing. The test value can be picked from any value within an interval between critical points or end points.
The First Derivative Test states:
- If \(f'(x) > 0\), \(f(x)\) increases (or has a positive slope)
- If \(f'(x) < 0\) \(f(x)\) decreases (or has a negative slope)
We can then identify the Max or Min points if there is a sign change between intervals. If it changes from positive to negative, then that point is a Max. If it changes from negative to positive, then that point is a Min. If there is no sign change between intervals, then that point is neither.
Example
For the function \(f(x) = 2x^3 + 3x^2 -36x +5\):
- Find the intervals of increase and decrease.
- Graph \(f(x)\) and \(f'(x)\) and explain how they indicate intervals of increase and decrease.
i. First, we can differentitate the function:
\(f(x) = 2x^3 + 3x^2 -36x +5\)
\(f'(x) = (3)(2x^2) + (2)(3x) -(1)(36) + (0)(5)\)
\(f'(x) = 6x^2 + 6x -36\)
Next, we can factor the derivative to solve for its critical numbers:
\(f'(x) = 6(x^2 + x -6)\)
\(f'(x) = 6(x + 3)(x-2)\)
\(x_1 = -3\)
\(x_2 = 2\)
After, we can conduct the First Derivative Test to determine the increasing and decreasing intervals:
Interval |
Test \(x\)-value |
\(f'(x)\) |
Conclusion |
\(\textcolor{green}{(-∞,-3)}\) |
\(-4\) |
\(f'(-4)=36\) |
\(f\) is Increasing |
\(\textcolor{red}{(-3,2)}\) |
\(0\) |
\(f'(0)=-36\) |
\(f\) is Decreasing |
\(\textcolor{green}{(2,∞)}\) |
\(3\) |
\(f'(3) = 36\) |
\(f\) is Increasing |
Therefore, we can determine that the Intervals of Increase are:
\(\textcolor{green}{x∈(-∞,-3)}\)
\(\text{AND}\)
\(\textcolor{green}{x∈(2,∞)}\)
We can also determine that the Interval of Decrease is:
\(\textcolor{red}{x∈(-3,2)}\)
ii. We can graph \(f(x)\) as such:
From this graph, we can determine:
- \(f(x)\) moving upward with respect to \(x\) means that \(f(x)\) is increasing
- \(f(x)\) moving downward with respect to \(x\) means that \(f(x)\) is decreasing
We can graph \(f'(x)\) as such:
From this graph, we can identify:
- \(f'(x)\) above the \(x\)-axis means that \(f(x)\) is increasing
- \(f'(x)\) below the \(x\)-axis means that \(f(x)\) is decreasing
Find the intervals of increase and decrease for the function \(f(x) = \cfrac{6-2x}{x^2-4}\).
Show Answer
First, we can determine \(f(x)\) and \(g(x)\) and their respective derivatives:
\(\textcolor{red}{f(x) = 6-2x}\)
\(\textcolor{red}{f'(x) -2}\)
\(\textcolor{blue}{g(x) = x^2-4}\)
\(\textcolor{blue}{g'(x) = 2x}\)
Next, we can implement the Quotient Rule to differentiate this expression:
\(f'(x) = \cfrac{\textcolor{blue}{g(x)}\textcolor{red}{f(x)'} - \textcolor{red}{f(x)}\textcolor{blue}{g(x)'}}{\textcolor{blue}{[g(x)]^2}}\)
\(f'(x) = \cfrac{\textcolor{blue}{(x^2-4)}\textcolor{red}{-2} - \textcolor{red}{(6-2x)}\textcolor{blue}{2x}}{\textcolor{blue}{(x^2-4)^2}}\)
\(f'(x) = \cfrac{-2(x^2-4) - 2x(6-2x)}{[(x+2)(x-2)]^2}\)
\(f'(x) = \cfrac{-2x^2+8 - 12x + 4x^2}{[(x+2)(x-2)]^2}\)
\(f'(x) = \cfrac{2x^2-12x+8}{[(x+2)(x-2)]^2}\)
Next, we can take a common factor out of the numerator:
\(f'(x) = \cfrac{2(x^2-6x+4)}{[(x+2)(x-2)]^2}\)
Next, we can determine the critical numbers by using the Quadratic Formula:
\(x = \cfrac{-\textcolor{blue}{b}\pm\sqrt{\textcolor{green}{b}^2-4\textcolor{red}{a}\textcolor{blue}{c}}}{2\textcolor{red}{a}}\)
\(x = \cfrac{-(\textcolor{green}{-6})\pm \sqrt{(\textcolor{green}{-6})^2-4(\textcolor{red}{1})(\textcolor{blue}{4})}}{2(\textcolor{red}{1})}\)
\(x = \cfrac{6\pm \sqrt{36-16}}{2}\)
\(x = \cfrac{6\pm \sqrt{20}}{2}\)
\(x = \cfrac{6\pm 2\sqrt{5}}{2}\)
\(x = 3\pm \sqrt{5}\)
Using \(\pm\), we can determine the 2 critical numbers, either by adding or subtracting the 2 values:
\(x_1 = 3 + \sqrt{5}\)
\(x_1 = 5.2\)
\(x_2 = 3 - \sqrt{5}\)
\(x_2 = 0.8\)
Aside from these values, we can also identify \(-2\) and \(2\) as critical numbers from the denominator of the derivative.
Finally, we can use the First Derivatives Test to determine the intervals of increase and decrease:
Interval |
Test \(x\)-value |
\(f'(x)\) |
Conclusion |
\(\textcolor{green}{(-∞,-2)}\) |
\(-1\) |
\(f'(-1)= 22/9\) |
\(f\) is Increasing |
\(\textcolor{green}{(-2,0.8)}\) |
\(0\) |
\(f'(0)=1/2\) |
\(f\) is Increasing |
\(\textcolor{red}{(0.8,2)}\) |
\(1\) |
\(f'(1)=-2/9\) |
\(f\) is Decreasing |
\(\textcolor{red}{(2,5.2)}\) |
\(3\) |
\(f'(3)=-13/18\) |
\(f\) is Decreasing |
\(\textcolor{green}{(5.2,∞)}\) |
\(6\) |
\(f'(6) = 1/128\) |
\(f\) is Increasing |
Therefore, we can determine that the Intervals of Increase are:
\(\textcolor{green}{x∈(-∞,-2)}\)
\(\text{AND}\)
\(\textcolor{green}{x∈(-2,0.8)}\)
\(\text{AND}\)
\(\textcolor{green}{x∈(5.2,∞)}\)
We can also determine that the Intervals of Decrease are:
\(\textcolor{red}{x∈(0.8,2)}\)
\(\text{AND}\)
\(\textcolor{red}{x∈(2,5.2)}\)