
How do you find the range of a function algebraically $y = \dfrac{{x + 5}}{{x - 2}}$?
Answer
558k+ views
Hint: An algebraic function is a function that involves only algebraic operations, like, addition, subtraction, multiplication, and division, as well as fractional or rational exponents. In the given algebraic equation we can observe that it is not valid for a particular value that is 2. When we give input as two the denominator will become zero which is not defined.
Complete step by step solution:
As given in the question, $y = \dfrac{{x + 5}}{{x - 2}}$
Now, to find the range of a function just replace x with y and y with x, and form a function of y instead of x, as given in the question,
So, after replacing x and y we will get our new equation as
$x = \dfrac{{y + 5}}{{y - 2}}$ ,
Now, multiply both sides with $y - 2$, so we will get
$(y - 2)x = y + 5$
$ \Rightarrow $ $xy - 2x = y + 5$
$ \Rightarrow $ $xy - y = 2x + 5$ (Taking y to left side and 2x to right)
$ \Rightarrow $$y(x - 1) = 2x + 5$ (Taking y as common in left hand side)
$ \Rightarrow y = \dfrac{{2x + 5}}{{x - 1}}$, now this is our new equation
And, the domain of new equation is $( - \infty ,1) \cup (1,\infty )$, which will be the range of ore original equation
Hence, our answer for this question is $( - \infty ,1) \cup (1,\infty )$.
Note: There will be a point in its graph of our new function where the function will not be defined, which is $x = 1$. If we input $x = 1$, the denominator will give zero which is not defined. One shortcut in this question is to just divide the coefficient of x in the numerator with the coefficient of x in the denominator and subtract it with the complete range, which is from negative infinity to positive infinity. This shortcut is only valid if the coefficient of x in numerator and denominator is equal.
Complete step by step solution:
As given in the question, $y = \dfrac{{x + 5}}{{x - 2}}$
Now, to find the range of a function just replace x with y and y with x, and form a function of y instead of x, as given in the question,
So, after replacing x and y we will get our new equation as
$x = \dfrac{{y + 5}}{{y - 2}}$ ,
Now, multiply both sides with $y - 2$, so we will get
$(y - 2)x = y + 5$
$ \Rightarrow $ $xy - 2x = y + 5$
$ \Rightarrow $ $xy - y = 2x + 5$ (Taking y to left side and 2x to right)
$ \Rightarrow $$y(x - 1) = 2x + 5$ (Taking y as common in left hand side)
$ \Rightarrow y = \dfrac{{2x + 5}}{{x - 1}}$, now this is our new equation
And, the domain of new equation is $( - \infty ,1) \cup (1,\infty )$, which will be the range of ore original equation
Hence, our answer for this question is $( - \infty ,1) \cup (1,\infty )$.
Note: There will be a point in its graph of our new function where the function will not be defined, which is $x = 1$. If we input $x = 1$, the denominator will give zero which is not defined. One shortcut in this question is to just divide the coefficient of x in the numerator with the coefficient of x in the denominator and subtract it with the complete range, which is from negative infinity to positive infinity. This shortcut is only valid if the coefficient of x in numerator and denominator is equal.
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