How to find the function value area?
The function can be constructed by points: substitute the value of a variable in the formula and put the corresponding points on the graph. But while there is no guarantee that you will not miss the point of extremum or gap. And the process is long and tedious. Therefore, it is much more rational to find the domain of definition, the domain of values and all the critical points of the function. Let's talk about this in more detail.
What is the scope of the function
The domain of the function y = f (x) is the set of all values of the function, which it takes when iterating through all values of x from the domain of definition x € X. Denotes the domain of the value as Е y = f (x).
About the scope is written in the article How to find the scope of a function. These two areas are sometimes confused, which is unacceptable. To better understand what it is, consider specific examples.
For example, the function y = f (x) = sinx. For clarity, you can draw a sine wave. Then we will see that x can vary from -∞ to + ∞, y = f (x) defined for x € -∞; + ∞. In this case, f (x) varies from -1 to +1, it does not accept other values. Hence, the domain of definition of the function x € -∞; + ∞, the range of values of E y = -1; +1 Those.The domain is the value of x for which the function exists. And the range of values is the values of the function that it takes in the whole range of definition.
Consider another simple example: y = 1 / x. We can also draw hyperbolas and know that for x = 0 the value of the function is not defined, i.e. at this point it does not exist. For x = 0, we have a discontinuity function. Hence, the domain of definition is x € (-∞ <0; 0 <∞), the range of the value of Е у = (-∞ <0; 0 <∞).
If we know the domain of the function, we need to find the maximum and minimum value of the function - this will be the range of values.
How to find the range of a function: an example
- We have the function y = 1 / (x² - 4).
We first look for the derivative of the function to find the extremum points.
- y '= (1 / (x² - 4))' = -2x / (x² - 4) ².
From this expression it follows that the first extremum point at x = 0, since at this point, the derivative changes sign. Since the sign changes from + to - this is the maximum.
The maximum value of the function at x = 0:
- y = 1 / (x² - 4) = y = 1 / (0² - 4) = -1 / 4.
- y max = -1/4.
Now we find the points of discontinuity of the function, which occur when the denominator of the derivative is 0.
- (x² - 4) ² = 0.
We decompose the expression into factors:
- (x - 2) (x + 2) = 0
The roots of the equation: x = 2; -2 This means that the points of discontinuity of the function. Determine what the function seeks at these points.
- Lim (1 / (x² - 4)) = lim1 (1 / (x - 2) (x + 2)) = lim (1 / (2 - 2) (2 + 2)) = lim ((1/0) (-1/4)) = -∞.
- x → - + 2
At the points of discontinuity, the function tends to minus infinity:
- With x = + -2 y = 1 / (x² - 4) → - ∞
It means that in the interval x = (-2; 0) y increases from -∞ to -1/4, and in the interval x = (0; 2) y decreases from -1/4 to ∞. Function area:
- E y = (-∞; -1/4).
General algorithm for determining the range of functions
- We take the derivative of the function in order to find the critical points: maximum, minimum, break points.
- Find the value of the function at the extremum points.
- Find the value of the limits of the function at the points of discontinuity.
- We define the range of the function. It is easier to do on the chart.
But if there is no time, you can also, it is easy and fast.
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