G(x)= sqrt. 5-X Find Domain and show in Interval Notation?
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, +∞).
Find the domain of the function f(x)= sqrt{x^2 - 81} in interval notation?
Domain of a function is/are that value(s) of "x" for which the function exists that is it gives result containing some definite value and not infinity. The given function is f(x)= sqrt{x^2 - 81} .....={x^2 - 9^2}^1/2 .....={(x+3)(x-3)}}^1/2 In this equation we may assign any value to x except +infinity and -infinity. Hence the domain is set of all real numbers excluding +infinity and -infinity.
What is the interval notation for all real numbers?
On the one hand, I am glad that interval notation is taught in American high schools. Far too little set theory is included in the curriculum, and the addition of notation to denote sets constitutes a small step to improve this. But, if we’re going to start to include more set theory, why not teach the symbol [math]\mathbb R[/math], recognized by mathematicians the world over as notation for the set of real numbers? High school students are consistently asked for function domains. What is the domain of [math]f(x) = \frac1x[/math]? Knowing interval notation ONLY, they respond with [math](-\infty, 0)\cup(0, \infty)[/math] (hopefully — the lack of formal training in set theory leaves them with no understanding of “union”). The notation [math]\mathbb R\setminus\{0\}[/math] is more compact, and represents directly the thought that the domain of [math]f[/math] is all real numbers except zero. Moreover, interval notation for [math]\mathbb R[/math], [math](-\infty, \infty)[/math], gives a description for the set of real numbers using some punctuation and two entities which are most certainly NOT real numbers! This sad state of affairs leaves students hopelessly confused later in calculus when limits are introduced. We write things such as [math]\lim_{x\to a} f(x) = \infty[/math], and then tell them (quite rightly) that the limit doesn’t exist. Recall the limit definition: in order for this limit to exist, there must be a real number [math]L[/math] such that, given any [math]\epsilon > 0[/math], there is a [math]\delta > 0[/math] so that whenever [math]0 < |x - a| < \delta[/math], it follows that [math]|f(x) - L| < \epsilon[/math]. But [math]\infty[/math] is NOT a real number! We simultaneously encourage students to treat [math]\pm\infty[/math] as if they’re real numbers on the one hand (wrong) and something else on the other.
How do you write {1} in interval notation?
The notation [math]{1}[/math] denotes a set with a single element: the value 1. If I had to describe the same set of numbers using interval notation, I would go with [math][1,1][/math], i.e., the closed interval that both begins and ends at 1.
How do you say X cannot equal 5, in interval notation?
the original question is to find the domain of f(x)=sqrt of 4x-20 and then put the answer in interval notation, im pretty sure the answer is all real numbers except 5
Find domain of f(x)=(sqrt(x+5))/(x+2). Express answer in interval notation.?
top is sqrt so must be +ve i.e. x+5> or = 0 x> or = -5 and bottom cannot equal 0 so x+2 (not=) 0 x (not=) -2 so we have x > or = 5, (not=)2
Differentiate f and find the domain of f?
1) Domain: You cannot take the ln of a non-positive number, so x-9 > 0 Therefore x > 9. In addition the denominator cannot be 0, but that would happen when ln(x-9) = 1 In that case x - 9 = e and x = 9 + e So: {x | x ix real, x > 9, and x ≠ 9 + e } <--------- Domain ----------------------- Differentiation Use the quotient rule: d(u/v)/dx = [(v*du/dx) - (u*dv/dx)] / (v^2) u = x and du/dx = 1 v = (1 - ln(x-9)) ........ dv/dx = (-1 / (x-9)) * 1 = (1 / (9 - x)) (I multiplied numerator and denominator by -1.) v^2 = 1 - (2*(ln(x-9))) + ((ln(x-9))^2) Now just substitute this into the quotient rule: ((1 - ln(x-9)) * 1) - ((x * (1 / (9 - x))) ---------------------------------------... <------- Answer 1 - (2*(ln(x-9))) + ((ln(x-9))^2) With (x-9) in numerators and denominators you can't simplify this. .
Find the domain of the logarithmic function f(x)=[(x+5)/(x-7)]?
I assume you're dealing only with reals. f(x) = log((x + 5)/(x - 7)) We need (x + 5)/(x - 7) > 0 and x - 7 ≠ 0. So x ≠ 7. i. x + 5 > 0 and x - 7 > 0, x > -5 and x > 7, x > 7. ii. x + 5 < 0 and x - 7 < 0, x < -5 and x < 7, x < -5. So the domain is (-∞,-5)∪(7,∞). f(x) = log(x^2 - 17x + 72). We need x^2 - 17x + 72 > 0, x^2 - 8x - 9x + 72 > 0, x(x - 8) - 9(x - 8) > 0, (x - 8)(x - 9) > 0. i. x - 8 > 0 and x - 9 > 0, x > 8 and x > 9, x > 9. ii. x - 8 < 0 and x - 9 < 0, x < 8 and x < 9, x < 8. So the domain is (-∞,8)∪(9,∞).