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Question: How do you find the range given \(x={{t}^{2}}-4\) and \(y=\dfrac{t}{2}\) for \(-2\le t\le 3\) ?...

How do you find the range given x=t24x={{t}^{2}}-4 and y=t2y=\dfrac{t}{2} for 2t3-2\le t\le 3 ?

Explanation

Solution

In order to find the range of the given function, we must first solve the inequality of the variable tt. By our prior knowledge of the domain and range of a function, we see that the range of the function is given by the variable yy. We shall then find the value of variable yy which represents the range of the function by substituting the appropriate values of tt obtained before.

Complete step by step answer:
If we have a function, let say ff , and if we give it a valid input of variable xx, then this function is going to map that to an output which we would call f(x)f\left( x \right). This output is also represented as the variable yy.

Thus, the range of the function is the set of all possible outputs that the function can produce.
Since, we know that yy gives the range of the function, therefore, we must find an interval of values ofyy.
Given that, y=t2y=\dfrac{t}{2}, thus we shall find the value of t2\dfrac{t}{2} first as it is equal to yy.
Also, given that 2t3-2\le t\le 3, so we will divide this entire inequality by 22.
22t232\Rightarrow \dfrac{-2}{2}\le \dfrac{t}{2}\le \dfrac{3}{2}
1t232\Rightarrow -1\le \dfrac{t}{2}\le \dfrac{3}{2}
This implies that 1y32-1\le y\le \dfrac{3}{2} becausey=t2y=\dfrac{t}{2}.
Therefore, the range of the given function is [1,32]\left[ -1,\dfrac{3}{2} \right].

Note: A domain is the set of all of the inputs over which the function is defined. If we input a value xx from this domain then the function will output another value, f(x)f\left( x \right). However, if we put a value xx which is not in the domain then the function would not be able to give a definite value.