Question Video: Finding the Domain of a Square Root Function Involving Absolute Values Mathematics

Consider the function 𝑓(π‘₯) = √(4 βˆ’ |π‘₯ βˆ’ 5|). 1. Find the domain of 𝑓(π‘₯). 2. Find the range of 𝑓(π‘₯).

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Video Transcript

Consider the function 𝑓 of π‘₯ equals the square root of four minus the absolute value of π‘₯ minus five. Part one, find the domain of 𝑓 of π‘₯, and part two, find the range of 𝑓 of π‘₯.

In this problem, we have a composite square root function of the form 𝑓 of π‘₯ equals the square root of 𝑔 of π‘₯. We recall that the domain of a composite square root function is the set of all values of π‘₯ for which 𝑔 of π‘₯ is nonnegative. For this function, this gives the inequality four minus the absolute value of π‘₯ minus five is greater than or equal to zero.

To solve this inequality, we first need to isolate the absolute value function. We can subtract four from each side of the inequality and then multiply or divide each side of the inequality by negative one, remembering that when we do we need to reverse the direction of the inequality sign. So we have the absolute value of π‘₯ minus five is less than or equal to four. If the absolute value of π‘₯ minus five is less than or equal to four, this means that the distance of the expression π‘₯ minus five from zero must be no greater than four. Or in other words, π‘₯ minus five is greater than or equal to negative four and less than or equal to four.

To solve this double-sided inequality, we add five to each part, giving π‘₯ is greater than or equal to one and less than or equal to nine. So we found that the domain of the function 𝑓 of π‘₯ is the closed interval from one to nine.

Now, let’s consider the range of 𝑓 of π‘₯. As the range of a function is the set of all possible values of that function, we can obtain the range by considering what the largest and smallest values the function can take are. As our function 𝑓 of π‘₯ is a square root function, we know that it cannot output negative values. Therefore, as a minimum, 𝑓 of π‘₯ is greater than or equal to zero. Zero would be the smallest value of the function if it is possible to produce zero from a value in the domain of 𝑓 of π‘₯.

To see whether zero is indeed a possible output value for this function, we would need the expression under the square root to be equal to zero because the square root of zero is zero. This would give the equation four minus the absolute value of π‘₯ minus five is equal to zero. Solving this absolute value equation gives the absolute value of π‘₯ minus five equals four. This means π‘₯ minus five is either equal to negative four or four, leading to π‘₯ equals one or π‘₯ equals nine. Both of these values are in the domain of the function 𝑓 of π‘₯, which means it is possible to achieve zero using a value in the domain.

So we’ve found the smallest possible value of 𝑓 of π‘₯. And now let’s consider what the largest possible value is. The largest value of 𝑓 of π‘₯ will correspond to the largest value of 𝑔 of π‘₯, which will in turn correspond to the smallest possible value of the absolute value of π‘₯ minus five. The absolute value function is always nonnegative, so its smallest value occurs when it is equal to zero. This occurs when π‘₯ is equal to five, which is in the domain of our function.

At this point, 𝑔 of π‘₯ will be four minus zero, which is four, and 𝑓 of π‘₯, remember, is the square root of 𝑔 of π‘₯. So it’s the square root of four, which is two. The largest value of 𝑓 of π‘₯ then is two. As 𝑓 of π‘₯ is a continuous function, its range will be everything from its smallest value to its largest value. So it is the closed interval from zero to two.

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