domain and range?

Question

Find the domain and range of f (x) = sq. root x + 5/x - 1

x + 5/x - 1 is under one square root.

..........

ok this is what i mean

____
/x + 5
/-------
√ x - 1

Answer 1

Domain (D):
x can be any real number except any x that causes the radicand (expression inside the radical symbol) to be negative. Also, any x that causes division by zero is out of the domain. So, we can immediately throw out x = 1.

x + 5
――– ≥ 0
x - 1

So, how do you figure out when the fraction is positive or negative? One way is to make a chart. Values of interest include x = -5 and x = 1. Let's look at the entire real number line...

-∞ ←――――――― -5 ―――――1 ―――――→ +∞

x+5→- - - - - - - - - - - - 0 + + + + + + + + + + + + +

x-1 →- - - - - - - - - - - - - - - - - - - - - - 0+ + + + + + +

x + 5
――– →+ + + + + + + 0 - - - - - - - - ∞+ + + + + + +
x - 1

Therefore, it appears that your radicand is non-negative from negative infinity up to and including -5. Also, from 1 to positive infinity the radicand is non-negative. However, as mentioned earlier, bad things happen at +1 so we'll exclude +1. So, in interval notation, your domain is given by the following:
D = (-∞, -5] U (1, ∞).

Range (R):
When you take the square root of a non-negative number, you get a non-negative number. Notice that your radicand will never equal 1. There is a rule for determining this. It has something to do with (ax+b)/(cx+d)→a/c as x→±∞. You'll need to look that up in your textbook or discover it on your own by thinking about it.

Therefore, R = [0, 1) U (1, +∞).

Answer 2

Quick note: I'm assuming that you mean exactly what you write: namely, that the expression under the square root sign is

x + (5/x) - 1

If you really meant

(x+5)(x-1)

then the previous answer probably will be more helpful. :)

The domain of a function is the set of all values that can be used for the variable -- think of it as the set of "allowable inputs." In this case, we can say x=0 is not in the domain, because that would give us zero in the denominator of a fraction. Beyond that, the value under the square root sign needs to be non-negative, which means we can eliminate all x < 0, because if x < 0, all three terms are negative, giving us a negative result. Beyond that, we need to solve the inequality:

x + 5/x - 1 ≥ 0

Since we're assuming x ≠ 0, we can multiply both sides by x:

x² + 5 - x ≥ 0

Rewriting the terms in order,

x² - x + 5 ≥ 0

Now, x² - x + 5 is a parabola, and in order to find the values of x that makes the expression ≥ 0, we need to find its intercepts -- in other words, solve the equation x² - x + 5 = 0. Problem is, x² - x + 5 has no real solutions, and therefore no x-intercepts. So the parabola is either entirely above the x-axis or entirely below it -- and since it opens upwards, it must be entirely above it. That means that any positive value of x makes that expression positive. (That may seem obvious, but it's useful to be able to prove it.)

So, the domain of the function is x > 0. What about the range? The output values can become infinitely large, and we know that they can't become negative, but how close to zero can they get on the positive side? That takes some calculus. We want to find the minimum value of sqrt(x + 5/x - 1) = (x + 5x^-1 - 1)^1/2. We differentiate and find the value(s) of x where the derivative becomes zero, which will be the minimum point(s), and then plug those x-values back into the original function.

f(x) = (x + 5x^-1 - 1)^1/2
f'(x) = (1/2)(x + 5x^-1 - 1)^(-1/2)(1 - 5x^-2) = [1 - 5/x²] / [ 2(x + 5/x - 1)^(1/2) ]

So to find the values where f'(x) = 0, we solve 1 - 5/x² = 0, getting x = +/- sqrt(5). Since we're limiting ourselves to positive values of x, that makes sqrt(5) the x-value at the minimum point. Plugging sqrt(5) into the original function:

f(sqrt(5)) = sqrt( 2√5 -1 ) ≈ 1.863367

A quick test of nearby values on the calculator seems to confirm this as a minimum. So the range of the function is y >= sqrt( 2√5 -1 ).

Hope that helps!

Answer 3

First off the proper way to write it is to include brackets

i.e., f(x) = sqrt ( x+5/x - 1)

Clearly x is not 0.

The sqrt is defined (i.e., real) if x+5/x-1>=0

So we need to solve the inequality x+5/x-1>=0

if x > 0 then we can multiply both sides by x and preserve the inequality

so; x^2-x+5 >=0

if x < 0 then we multiply both sides by x and reverse the inequality

so; x^2-x+5 <=0

We can notice that x^2 - x + 5 = ( x - 1/2) ^2 + (5-1/4)

Thus this is always positive.

So there are no solutions for x < 0

for x > 0 the inequality is always satisfied.

Thus our domain which is all valid x for f(x); is x > 0.

For the range we need to find the values of f(x) for x > 0.

Let g(x) = x + 5/x - 1

Then g'(x) = 1-5/x^2; Solving g'(x) = 0 gives x = sqrt(5)

Thus the minimum of g(x) is g(sqrt(5)) = 2*sqrt(5)-1 (no point in verifying it is indeed a minimum); and since g(x) is continuous on (0,infty) and is unbounded as x -> 0. Then g(x) has range (2*sqrt(5)-1,infty)

Thus f(x) has range (sqrt(2*sqrt(5)-1),infty)

Answer 4

The 'domain' of a function is the set of values that the variable can assume. In this case, x > 0 works because if x = 0 then the 5/x term is undefined and, if x < 0, you'd have to take the square root of a negative value which is usually not allowed (unless you're in a graduate level course in complex analysis)

The 'range' of a function is the set of values the function can produce for any variable from it's domain. In this case 0

Doug

Answer 5

Domain 00 but not equal to zero
Range y >0

Answer 6

domain is xis not equal to 0 and x is non negative
infinity>x>0
range all positive nos