Find domain equation, graph function interval notation, (f-g)(3)?
1) f(x) is well defined except where 4x²-5x+1 = 0 => (4x-1)(x-1)=0 => x = 1/4 or x = 1 So the domain is the set of real numbers except 1/4 and 1. You can write that in set notation as (-∞,1/4) ∪ (1/4,1) ∪ (1,∞) 2) If y = (x+5)/2 then x+5 = 2y x = 2y-5 so to get x from y we need f^(-1)(y) = 2y-5 and changing the names of the variable f^(-1)(x) = 2x-5 3) f(x) is well-defined for all real numbers. So the domain in interval notation is (-∞,+∞) 4) Not sure what you mean here. Do you really mean f-g? Or is it composition of functions?
Find the domain of f. Write answer using interval notation. F(x)=12x/x^2+4?
Domain is all of the x-values that will give you a y-value when subbed into the function. Now, if the denominator of a fraction is 0, we won't get an x-value, because you can't divide a number by 0. So you need to solve for: x^2+4=0 Now, we know that any number squared will be positive. Furthermore, no positive number + 4 will equal 0, it will always be a positive number. There is no value that would make the denominator 0 in the given function, meaning every single x-value works, making the domain all real numbers. All real numbers can be written in interval notation as (-infinity, infinity), because every value works. A brief description of interval notation: Interval notation can be used to write the domain or range of a function. The outside will either have brackets or parentheses, brackets are used if the domain/range includes the number on the given side, while parentheses are used when the number is not included in the domain/range. For example, [-3,3] means that the domain/range goes from -3 to 3 and includes those two values. (-3,3) means that the domain/range goes from -3 to 3 but doesn't include -3 and 3 themselves. If you have something like [-3,3), it means that the domain/range goes from -3 to 3 and includes -3 but not 3.
How do I find the average slope of a function at the interval (-∞, ∞)? i.e. a quadratic or cubic function?
Say the function is [math]f(x)[/math]. Find it's derivative [math]f'(x)[/math]The average of any function [math]g(x)[/math] on interval [math](a,b)[/math] is [math]\frac{1}{b-a} \int_a^b g(x) dx[/math]So assuming that the derivative is continuous on the domain of [math](-\infty,\infty)[/math], and [math]a=b=t\rightarrow \infty[/math] the average slope is [math]avg=lim_{t \to \infty} \frac{1}{2t} \int_{-t}^t f'(x) dx=[/math] [math]lim_{t \to \infty}[/math][math] \frac{f(t)-f(-t)}{2t}[/math]
How do I determine the intervals of the increase of a function?
thanks for A2A Anubhav Singhfind the first derivative of function m=f '(x) that tell you the slop of that function. now take [a,b] interval put the value in first derivative.m < 0 decreasingm = 0 constantm > 0 increasing particularly at that point. so for example f(x) = x^2-2xtake three or four value for x like -2, -1, 0, 1, 2 and plot it.Without exact analysis we cannot pinpoint where the curve turns from decreasing to increasing, so let us just say:Within the interval [-2,2]:the curve decreases in the interval [-2, 1]the curve increases in the interval [1, 2]2. f(x) = e^x this function is increasing in any interval.I tried my best.hope :)
Interval Notation?
The domain is all numbers except when the denominator is zero. This happens when x^2 - 4 x - 21 = 0 (x - 7) (x + 3) = 0 x = 7 or x = -3 so the domain is IR - {-3, 7}, or (-oo, -3) U (-3, 7) U (7, +oo). The range is harder to find. The denominator is a quadratic function with positive coefficient for x^2, and therefore has a minimum value. It takes on negative values (e.g. it is -21 when x = 0), and therefore the denominator can have all positive values. Therefore the range also includes the positive numbers. The minimum value of the denominator is found when x = 2 (the average of the zero points, -3 and 7); it is -25. Therefore the function f has maximum negative value 5/(-25) = -1/5. Conclusion: range is (-oo, -1/5] U (0, +oo).
Help finding the range of the quadratic function -x^2 + 2x + 1?
quadratic formula ax^2+bx+c a=-1,b=2,c=1 find vertex by using rule x= -b/2a x=-2/2.(-1)=-2/-2=1 find y =-(1)^2+2(1)+1=-1+2+1=2 so coordinates of vertex=(1,2) parabola concave down since a is negative( -) so range (- infinity,2 closed from right)
Graph each function, state it's domain and range: f(x)=4?
Sandra Need a little more information You indicate a f(x) = 4, then I think you are trying to say the function is x = (y + 2) ^2 Then is seems as if there is more than one function. for the function given x = (y + 2) ^2, you would typically solve for y to get y = √(x) - 2 then the domain would be the set of all real number greater than or equal to 0, and the range would be -2 to positive infinity the y intercept would be the point (0, -2) and the x intercept would be ( 4,0) the graph will start at point (0,-2) but will not have an end point Math is fun has a grapher that will graph the function for you, input the formula as x^(1/2) - 2. As for a vertex, since this equation is not parabolic or a higher order polynomial, not sure what they are looking for. Hope that helps, and you may want to post additional information for additional assistance.
A function f is given by the equation [math]f(x)=\dfrac{x^2+3x-10}{x-2}[/math]. How do you find the function's range?
This question just showed up on my feed, and I just wanted to take a shot at it for fun.I see both algebra and calculus listed in the topics. Since no one has used limits, here’s a slightly different approach. Nothing too fancy.[math]f(2)[/math] is undefined. That’s obvious, its a rational function.[math]\displaystyle \lim_{x\to 2} f(x)=\lim_{x\to 2} \dfrac{x^2+3x-10}{x-2}[/math][math]=\displaystyle \lim_{x\to 2} \dfrac{(x+5)(x-2)}{x-2}[/math][math]=\displaystyle \lim_{x\to 2} x+5[/math][math]=7[/math]This function [math]f(x)[/math] is discontinuous at [math]x=2[/math] and the discontinuity is removable.If we make the function piece-wise[math]f(x)=\begin{cases} \dfrac{x^2+3x-10}{x-2}, & x\neq 2 \\ 7, & x=2\end{cases}[/math]Then we see that we have been able to define the function and hence make it continuous at [math]x=2[/math].Using the idea of continuity, we can say that the range of this function is [math]\R-\{7\}[/math]I know this answer may get tons of negative comments, but I simply tried to work a little differently.
Functions and Quadratics Help Please Need Explaination?
1.) f(x) = 4x² - 1 g(x) = 16x + 47 Points of Intersection: f(x) = g(x): 4x² - 1 = 16x + 47 4x² - 16x - 1 - 47 = 0 4x² - 16x - 48 = 0 4(x² - 4x - 12) = 0 x² - 4x - 12 = 0 (x + 2)(x - 6) = 0 If the product of two factors equals zero, then one or both factors equal zero. If x + 2 = 0, x = - 2 and y = 16(- 2) + 47 y = - 32 + 47 y = 15 One Point of Intersection (- 2, 15) ........ Q2 If x - 6 = 0, x = 6 and y = 16(6) + 47 y = 96 + 47 y = 143 Other Point of Intersection (6, 143) ...... Q1 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 2.) Vertex (5, - 3): h = 5 k = - 3 x = - 2 y = 0 a.) y = a(x - h)² + k Subbing known values, 0 = a(- 2 - 5)² + (- 3) 0 = a(- 7)² - 3 0 = a(49) - 3 0 = 49a - 3 49a = 3 a = 3/49 Equation: y = 3/49 (x - 5)² - 3 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ b.) Range: [- 3, ∞) ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 3.) f(x) = - 4x² - 24x - 38 y = (- 4x² - 24x) - 38 y = - 4(x² + 6x) - 38 y = - 4(x² + 6x + 9) - 38 - [- 4(9)] y = - 4(x + 3)² - 38 - (- 36) y = - 4(x + 3)² - 38 + 36 y = - 4(x + 3)² - 2 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
How do you determine the domain and range of a logarithmic function?
The domain of a function is the set of inputs it can take *. The range of a function is the set of outputs it can produce from the inputs in the domain.The domain of [math]f(x) = \log(9 - x^2)[/math] can be checked by decomposing it into compositions **. Note that [math]f(x) = f_3(f_2(f_1(x)))[/math] where [math]f_1(x) = x^2, f_2(x) = 9 - x[/math], and [math]f_3(x) = \log x[/math].[math]f_3(x) = \log x \in \mathbb{R}[/math] only if [math]x \in (0, \infty)[/math].[math]f_2(x) = 9 - x \in (0, \infty)[/math] only if [math]x \in (-\infty, 9)[/math].[math]f_1(x) = x^2 \in (-\infty, 9)[/math] only if [math]x \in (-3, 3)[/math].Thus our domain is [math](-3, 3)[/math]. Now that we have the domain, we can determine the range in a similar reversed process.[math]f_1(x) = x^2[/math] maps [math](-3, 3)[/math] to [math][0, 9)[/math].[math]f_2(x) = 9 - x[/math] maps [math][0, 9)[/math] to [math](0, 9][/math].[math]f_3(x) = \log x[/math] maps [math](0, 9][/math] to [math](-\infty, \log 9][/math].Thus our range is [math](-\infty, \log 9][/math].* Technically, the domain can be any subset of the possible inputs. When a function is given without specifying the domain, a typical assumption is that the intended domain is the largest possible subset of the real numbers that makes sense. In other contexts, it may make more sense to consider domains that include complex numbers as well.** For a multiple choice question, it might be easier to just plug in some numbers to eliminate choices.