Increasing and Decreasing Functions
In the study of calculus, understanding how functions behave is crucial. One key aspect of this is identifying when a function is increasing or decreasing.
Defining Increasing and Decreasing Functions
A function $f(x)$ is said to be increasing on an interval if, for any two points $x_1$ and $x_2$ in that interval where $x_1 < x_2$, we have $f(x_1) < f(x_2)$. Conversely, a function is decreasing on an interval if, for any two points $x_1$ and $x_2$ in that interval where $x_1 < x_2$, we have $f(x_1) > f(x_2)$.
NoteIt's important to distinguish between strictly increasing/decreasing and non-strictly increasing/decreasing functions. The definitions above describe strictly increasing/decreasing functions. For non-strict versions, we would use $\leq$ and $\geq$ instead of $<$ and $>$.
The Role of Derivatives
The derivative of a function plays a crucial role in determining whether a function is increasing or decreasing.
- If $f'(x) > 0$ on an interval, then $f(x)$ is increasing on that interval.
- If $f'(x) < 0$ on an interval, then $f(x)$ is decreasing on that interval.
- If $f'(x) = 0$ at a point, the function may have a horizontal tangent line, a local maximum, a local minimum, or an inflection point at that point.
Consider the function $f(x) = x^3 - 3x^2 + 2$. Its derivative is $f'(x) = 3x^2 - 6x$.
To find the intervals where $f(x)$ is increasing or decreasing:
- Solve $f'(x) = 0$: $3x^2 - 6x = 0$ $3x(x - 2) = 0$ $x = 0$ or $x = 2$
- These points divide the real line into three intervals: $(-\infty, 0)$, $(0, 2)$, and $(2, \infty)$.
- Test a point in each interval:
For $x = -1$: $f'(-1) = 3(-1)^2 - 6(-1) = 9 > 0$
For $x = 1$: $f'(1) = 3(1)^2 - 6(1) = -3 < 0$
For $x = 3$: $f'(3) = 3(3)^2 - 6(3) = 9 > 0$
Therefore, $f(x)$ is increasing on $(-\infty, 0)$ and $(2, \infty)$, and decreasing on $(0, 2)$.
ExampleA company's profit function (in thousands of dollars) for producing $x$ units of a product is given by $P(x) = -0.01x^2 + 4x - 100$.
To find when the profit is increasing:
- Find $P'(x) = -0.02x + 4$
- Solve $P'(x) > 0$:
$-0.02x + 4 > 0$
$-0.02x > -4$
$x < 200$
Therefore, the profit is increasing for $0 < x < 200$ units.
Understanding increasing and decreasing functions is fundamental in calculus and forms the basis for more advanced topics like optimization and curve sketching. By mastering this concept, students gain valuable tools for analyzing function behavior in various contexts.
