Need to find a range of definition of function arises at the solution of any task on a research of its properties and creation of the schedule. Only on this set of values of an argument it makes sense to make calculations.
Instruction
1. To find a range of definition is the first that should be done during the work with functions. This set of numbers which possesses a function argument, with imposing of some restrictions following from use in its expression of certain mathematical designs, for example, of a square root, fraction, a logarithm, etc.
2. As a rule, types and their various combinations can refer all these structures to six main. It is necessary to solve one or several inequalities to define points in which function cannot exist.
3. Power function with an exponent in the form of fraction with an even znamenatelemeto function of a look u^(m/n). It is obvious that a radicand negative, therefore, it cannot be necessary to solve inequality of u≥0. Example 1: at = √ (2 • x - 10). Decision: make inequality 2 • x – 10 ≥ 0 → x ≥ 5. A range of definition - an interval [5; + ∞). At x
4. Logarithmic function of a look log_a (u) In this case inequality will be strict u> 0 as expression under the sign of a logarithm cannot be less than zero. Example 2: at =log_3 (x - 9). Decision: x – 9> 0 → x> 9 → (9; + ∞).
5. Fraction of a type of u (x)/v (x) it is obvious that a denominator of fraction to address in zero, so critical points cannot be found from equality of v (x) = 0. Example 3: at = 3 • x² – 3 / (x³ + 8). Decision: x³ + 8 = 0 → x³ =-8 → x =-2 → (-∞;-2) U (-2; + ∞).
6. Trigonometrical tg u and ctg unaydite functions of restriction from inequality of a look x ≠ π/2 + π\• k. Example 4: at = tg (x / 2). Decision: x / 2 ≠ π/2 + π\• k → x ≠ π\• (1 + 2•k).
7. Trigonometrical arcsin u and arccos ureshite functions bilateral inequality-1 ≤ u ≤ 1. Example 5: at = arcsin 4 • x. Decision:-1 ≤ 4 • x ≤ 1 →-1/4 ≤ x ≤ 1/4.
8. Indicative power functions of a type of u (x) ^v (x) the Range of definition has restriction in the form of u> 0. Example 6: at = (x³ + 125) ^sinх. Decision: x³ + 125> 0 → x>-5 → (-5; + ∞).
9. Presence at function of two or more of the given expressions assumes at once imposing of more strict restrictions considering all components. It is necessary to find them separately, and then to unite in one interval.