# How to find intervals of increasing functions

Suppose that a function - f (x), defined by its equation. The challenge is to find the gaps it monotonically increasing or monotonically decreasing.

## Instruction how to find gaps increasing functions

Step 1:

f (x) is called a monotonically increasing function over the interval (a, b), if for all x, belonging to this interval, f (a) lt; f (x) lt; f (b). The function is called monotonically decreasing on the interval (a, b), if, for any x, belonging to this interval, f (a) gt; f (x) gt; f (b). If not complied with any of these conditions, the function can not be called either monotonically increasing or monotonically decreasing. Additional research is needed in these cases.

Step 2:

The linear function f (x) = kx + b increases monotonically throughout its domain if k gt; 0, and monotonically decreasing if k lt; 0. If k = 0, the function is a constant, and it is neither increasing nor decreasing.

Step 3:

The exponential function f (x) = a ^ x increases monotonically throughout the domain of definition, if a gt; 1, and decreases monotonically if 0 lt; a lt; 1. If a = 1, the function, as in the previous case, is converted into a constant.

Step 4:

In general, the function f (x) may have a predetermined number of section intervals of increase and decrease. To find them, you need to explore it to extremes.

Step 5:

Given a function f (x), its derivative is denoted f '(x). The original function has an extremum point where its derivative vanishes. If the passage of this point of the derivative changes sign from plus to minus, then found the maximum point. If the derivative changes sign from minus to plus, is found extremum - minimum point.

Step 6:

Let f (x) = 3x ^ 2 - 4x + 16, and the interval at which it is necessary to investigate - (3, 10). The derivative of the function is f '(x) = 6x - 4. It vanishes at the point xm = 2/3. Since f '(x) lt; 0 for any x 0 for any x gt; 2/3, the found point of a function f (x) is minimum. Its value at this point is equal to f (xm) = 3 * (2/3) ^ 2 - 4 * (2/3) + 16 = 14, (6).

Step 7:

Discovered at least lies within the boundaries of the specified area. For further analysis is necessary to compute f (a) and f (b). In this case: f (a) = f (-3) = 3 * (- 3) ^ 2 - 4 * (- 3) + 16 = 55, f (b) = f (10) = 3 * 10 ^ 2 - 4 * 10 + 16 = 276.

Step 8:

Since f (a) gt; f (xm) lt; f (b), the given function f (x) decreases monotonically on the interval (-3 2/3) and monotonically increasing on the interval (2/3, 10).