The domain of the function: how to find it

The need to find the area of definition of the function arises in the solution of any problem to the study of its properties and graphing. Only on this set of values of the argument makes sense to perform the calculations.

Instruction

1. Find the area definition is the first thing to do when working with functions. This set of numbers, which belongs to the argument of the function, with the imposition of certain restrictions arising from the use of it in terms of certain mathematical structures, for example, square root, fractions, logarithms, etc.
2. As a rule, all these structures can be attributed to six main types and their various combinations. Need to solve one or more of the inequalities to determine the points at which a function cannot exist.
3. The power function with the exponent as a fraction with an even, znamenatelnih a function of the form u^(m/n). Obviously, radical expression cannot be negative, therefore, we need to solve the inequality u≥0.Example 1: y=√(2•x - 10).Solution: write the inequality 2•x – 10 ≥ 0 → x ≥ 5. Scope definition is the interval [5; +∞). When x
4. Logarithmic function of the form log_a (u)In this case, the inequality is strict, u>0, since the expression under the sign of the logarithm cannot be less than zero.Example 2: y=log_3 (x - 9).Solution: x – 9 > 0 → x > 9 → (9; +∞).
5. Fraction of the form u(x)/v(x)Obviously, the denominator cannot go to zero so critical points can be found from the equality v(x) = 0.Example 3: y = 3•x2 – 3/(X3 + 8).Solution: X3 + 8 = 0 → X3 = -8 → x=-2 → (-∞; -2)U(-2; +∞).
6. Trigonometric functions tg and ctg u inaudita constraints from the inequality x ≠ π/2 + π•k.Example 4: y = tg (x/2).Solution: x/2 ≠ π/2 + π•k → x ≠ π•(1 + 2•k).
7. Trigonometric functions arcsin u and asso onesite bilateral inequality -1 ≤ u ≤ 1.Example 5: y = 4 arcsin•x. Solution: -1 ≤ 4•x ≤ 1 → -1/4 ≤ x ≤ 1/4.
8. The exponential-power function of the form u(x)^v(x)Region definition is the restriction to u>0.Example 6: y = (X3 + 125)^p.Solution: X3 + 125 >0 → x > -5 → (-5; +∞).
9. The presence of functions of two or more of the above expressions imply the imposition of more stringent restrictions, taking into account all the components. You need to find them separately, and then combined into a single interval.