Find the domain and range for for following equations (a) -√(x^2+2x-3) (b)(x^2+x+1)/x

Answer 1

(a - domain) We know that this (x^2 + 2x - 3) is inside a square root, meaning that it can't be negative: [x^2 + 2x - 3 ≥ 0]. Therefore, what values of x make it so that the value inside the square root is nonnegative? If we set x^2 + 2x - 3 = 0 (just like a quadratic function), we can factor and get (x-1)(x+3) = 0. This means our zeroes (the values that make the equation 0) are 1 and -3, because (x-1) = 0 if x = 1, (x+3) = 0 if x = -3, and 0 x [anything] = 0. From here, we can make three intervals, and we can take any arbitrary number from each interval and plug it into the original equation.

1. Interval a (-∞, -3) → -5 lies between -3 and -∞, so let's plug in -5 (note that you could also plug in -4, -6, -7.5, etc, but I chose -5 because I thought it would be easiest). If we plug in 5, we get (5-1)(5+3) = (4)(8) = 32, which is positive. This means that if x equals any value within interval a, then the function [x^2 + 2x - 3] will be positive.
2. Interval b (-3, 1) → 0 lies between -3 and 1, so let's plug it in (again, you could choose other values within the interval as well). If we plug in 0, we get (0-1)(0+3) = (-1)(3) = -3, which is negative. This means that if x equals any value within interval b, then the function [x^2 + 2x - 3] will be negative.
3. Interval c (1, ∞) → 2 lies between 1 and ∞, so let's plug it in (again, you could choose other values within the interval as well). If we plug in 2, we get (2-1)(2+3) = (1)(5) = 5, which is positive. This means that if x equals any value within interval c, then the function [x^2 + 2x - 3] will be positive.

As we mentioned above, we know that the value of [x^2 + 2x - 3] must be nonnegative because it's inside a square root. Therefore, the only intervals in which this is true would be (-∞, -3] U [1, ∞) → note that i used inclusive brackets for the 1 and the -3 because these two values both make the equation equal to 0, which still works within a square root. Thus, the domain of (a) is (-∞, -3] U [1, ∞).

(a - range) Okay so now what do we know?

1. we know that the value inside the square root must be positive
2. we know that the square root of any value must be positive
3. we know that since we're taking the negative of a square root, our final value is going to be negative
4. we know that the quadratic [x^2 + 2x - 3] is a parabola facing upwards (you could easily graph it, or just notice that the coefficient in front of the x^2 term is positive so the values will go to infinity)

If we put all of this information together, we realize that the largest value that [-√(x^2 + 2x - 3)] can equal is 0.

1. x^2 + 2x - 3 = 0 (again, it cannot be negative because square root)
2. the square root of 0 is 0
3. the negative of 0 is 0

On the other hand, the smallest value that this expression can equal is -∞.

1. x^2 + 2x - 3 = ∞ (again, we found that the parabola goes up to ∞)
2. the square root of ∞ is ∞
3. the negative of ∞ is -∞

Thus, the range of this expression is (-∞, 0].

(b - domain) For the expression (x^2 + x + 1)/x, let's start out by looking at what x cannot be. In this case, since we have x in the denominator of a fraction, x cannot equal 0 because anything divided by 0 is undefined. This is really our only limitation, and we can see this if we actually divide the numerator by the denominator by x and get (x^2 + x + 1)/x = x + 1 + 1/x. The only issue here is the 1/x. Thus, our domain can equal anything and everything except for x = 0. We can represent this using interval notation and get (-∞, 0) U (0, ∞).

(b - range) There's a rule in math that states that if you have a function f(x) = y, you set the discriminant of the function to be ≥ 0 to find the range (the possible values of y represent the range of the function). In this case we have f(x) = (x^2 + x + 1)/x = y. If we algebraically manipulate the equation a bit, we get:

1. (x^2 + x + 1)/x = y
2. x^2 + x + 1 = xy
3. x^2 + x - xy + 1 = 0
4. x^2 + (1 - y)x + 1 = 0

The discriminant of a quadratic [ax^2 + bx + c = 0] is [b^2 - 4ac] → in this case, the discriminant is (1-y)^2 - 4(1)(1) = (1 - 2y + y^2) - 4 = y^2 - 2y - 3. If we set this as ≥ 0, we can solve it the same way we solved the inequality in (a - domain).

1. y^2 - 2y - 3 = 0
2. (y-3)(y+1) = 0 → y = 3, -1
3. 3 intervals: (-∞, -1), (-1, 3), (3, ∞)
4. (-∞, -1) → select -2: (-2 - 3)(-2 + 1) = (-5)(-1) = 5
5. (-1, 3) → select 0: (0-3)(0+1) = (-3)(1) = -3
6. (3, ∞) → select 4: (4-3)(4+1) = (1)(5) = 5

The discriminant is positive when y within the intervals (-∞, -1) and (3, ∞). Therefore, the range of the function is (-∞, -1] U [3, ∞), because these are the values of y that make the discriminant greater or equal to 0.

Answer 2

SOLUTION OF THE FIRST PROBLEM

ƒ( x) = -√(x^2+2x-3)

In order fur this function to be defined the quantity under the radical must be non negative.

That means we must solve the inequality x² + 2x - 3 ≥ 0.

Setting the zeros of the equation x= -3 and x = 1 and see how the sign of the quantity x² + 2x - 3

changes for various values of x we see that the quantity is greater or equal to zero when

x ∈ ( −∞, - 3 ] ∪ [ 1, ∞ )

Now the range of the function is ( −∞ ,0]

SOLUTION TO THE SECOND PROBLEM

ƒ( x )=(x^2+x+1)/x

For this one we can easily say that the only value of x for which the function is not defined is zero

That is the domain of the function is all real numbers except zero .

Finding the range though is little more complicated.

Set f(x) =y

What we are actually after for is for which values of y the equation ( x² + x + 1 )/ x = y when solved for x

has solutions.

That is find the values of the y's that make the equation x² + x + 1 =x y solvable.

Equivalently x² + ( 1 -y ) x +1 = 0

For the above equation to have real solution for x , its discriminant must be greater or equal to zero.

Therefore ( 1 - y )² - 4 ≥ 0 or y² - 2 y - 3 ≥ 0

Solving the above inequality for y we get y ∈ ( −∞ , −1] ∪ [ 3 , ∞ )