Find range of reciprocal or rational function?
The range of a function is the same as the domain of its inverse. y = 1/x + 1 To find the inverse, switch x and y, then solve for y. x = 1/y + 1 Subtract 1 from both sides. x-1 = 1/y Multiply both sides by y. y(x-1) = 1 Divide both sides by x-1. y = 1/(x-1) The domain is all real numbers except 1. That means the range of the original is all real numbers except 1.
What is a real life example of a rational/reciprocal function that I could capture in a photo?
Interesting and hardest question... One example is... Relationship between the length and the width of the picture you want to take... You can set the frame (with some interesting shape, like the diamond) inside your camera, so you can position the picture/frame in your camera. Good luck!
How do I manipulate the reciprocal function (f(x) =1/x), things such as stretching the graph vertically or reflection across the x-axis?
Let’s take a function [math]f(x)[/math], not necessarily the reciprocal function, and look what we can do to it.The function [math]g(x) = af(x)[/math] is the same function, but scaled vertically.The function g[math](x) = f(x) + a[/math] is the same function, but translated vertically.The function [math]g(x) = f(ax)[/math] is the same function, but scaled horizontally.The function [math]g(x) = f(x+a)[/math] is the same function, but translated horizontally.If the scaling factors are negative, it’ll be a reflection about that axis, so [math]g(x) = -f(x)[/math] is a reflection about the x-axis, and [math]g(x) = f(-x)[/math] is a reflection about the y-axis.You can combine these as well, so [math]g(x) = -5f(4x+3)+2[/math] is reflected about the x-axis, scaled both horizontally and vertically, and translated both horizontally and vertically.If you were dealing with an equation for a curve, like [math]f(x,y) = 0[/math], you could scale, reflect, and translate by basically doing the same things. But you can also rotate by using [math]g(x,y) = f(ax+by,bx-ay) = 0, a^2+b^2 = 1[/math] as another operation. Generally speaking, [math]g(x,y) = f(ax+by+c, rx+sy+t) = 0[/math] is a rotated, reflected, scaled, and translated version of [math]f(x,y) = 0[/math].
What is a quadratic function whose reciprocal has no vertical asymptotes?
The vertical asymptotes occur when the reciprocal approaches zero. A function that does not equal zero at any point will not have vertical asymptotes. For example, here is [math]f(x)=x^2+1[/math]:Here is it’s reciprocal: [math]g(x)=\frac{1}{x^2+1}[/math]:No vertical asymptotes here!
What is the difference between inverse and reciprocal functions?
Inverse means the negative or opposite of something. Reciprocal means that you flip the fraction around. Inverse of 5 = -5 Reciprocal of 5 = 1/5 They effect functions like... y = x + 2 Reciprocal of the slope makes it.... y = 1/x + 2 Inverse of the slope makes it.... y = -x + 2
Range and Domain of a reciprocal function? 10 points!?
If that's supposed to be 1/(-x^2 + 6), you need the parentheses. the / doesn't stretch over or under anything the way a horizontal fraction bar does, so it's normally used as a synonym for ÷, which groups like mulltiplication. (Remember PEMDAS?) I'll guess you meant the parenthesized version, though, because it would be more normal to write 1/-x² as -1/x². I'd still rewrite it as: y = 1/(6 - x²) That's only undefined when the denominator is 0...that is, when x² = 6. That happens in two places: x=sqrt(6) and x=-sqrt(6). Those two points break up the domain into three pieces: x < -sqrt(6); -sqrt(6) < x < sqrt(6); and x > sqrt(6) The union of those three pieces is your domain. The denominator can be any number less than or equal to 6, except for zero. The intervals for the denominator are: 0 < 6 - x² <= 6 ....OR.... -infinity < 6 - x² < 0 Taking reciprocals ( I put the -infinity "limit" in so you can take the reciprocal and get 0 here) gives: infinity > 1/(6 - x²) >= 1/6 ....OR.... 0 > 1/(6 - x²) > -infinity In more conventional terms, removing infinities, you have the two pieces: x >= 1/6 ....OR.... x < 0
If cotangent is the reciprocal function of tangent, then why isn't cosine the reciprocal of sine? And cosecant the reciprocal of secant?
You wonder why [math]\cot\left(x\right)[/math] is the reciprocal function of [math]\tan\left(x\right)[/math], but why isn't [math]\cos\left(x\right)[/math] the reciprocal of [math]\sin\left(x\right)[/math] and why [math]\csc\left(x\right)[/math] isn’t the reciprocal of [math]\sec\left(x\right)[/math]?Okay first let's get clear with the meaning of "co-" first. In Mathematics particularly angle, "co-" means complement of an angle. Hence a co-trig function of "co-f(x)" is just f[math]\left(\frac{\pi}{2} - x\right)[/math]. For example look at the co-trig functions below:[math]\cos\left(x\right) = \sin\left(\frac{\pi}{2} - x\right)[/math],[math]\cot\left(x\right) = \tan\left(\frac{\pi}{2} - x\right)[/math],[math]\csc\left(x\right) = \sec\left(\frac{\pi}{2} - x\right)[/math].[math]\tan\left(x\right) = \frac{\sin\left(x\right)}{\cos \left(x\right)} [/math]Therefore,[math]\cot\left(x\right) = \tan\left(\frac{\pi}{2} - x\right) [/math][math] = \frac{\sin\left(\frac{\pi}{2} - x\right)}{\cos\left(\frac{\pi}{2} - x\right)} [/math][math] = \frac{\cos\left(x\right)}{\sin \left(x\right)} [/math][math] = [/math]reciprocal of [math]\tan\left(x\right) [/math][math]\cot\left(x\right)[/math] being the reciprocal of [math]\tan\left(x\right)[/math] is merely just a coincidence. The main idea behind a co-trig is the complement of angles. Hope this helps!
What is the difference between a reciprocal and an inverse function? In French those terms seem switched.
In French (at least where I’ve been taught) :“La fonction inverse” (the “inverse” function) is the function 1/x.“L’inverse de la fonction” = “La réciproque de la fonction” is the reciprocal, e.g. the inverse by composition : the function such that when you compose it with the one whom it’s the reciprocal, you get the identity (the function f(x) = x).Ln(x) is the reciprocal of Exp(x), arcsin the one from sin, but Exp(-x) is the inverse of Exp(x).I don’t think they are switched, but in French, “Inverse” is ambiguous, while reciprocal (“réciproque”) has always the same meaning.I don’t know if in English we can use “the inverse of a function” to represent its reciprocal. But we do in French.
Math Help!?!!! ASAP average rate of change, reciprocal function?
1.) f(x) = 5x² + 13x - 6 Reciprocal: g(x) = 1 / (5x² + 13x - 6) is a fraction and, as such, the denominator cannot equal zero, lest division by zero occurs, which is undefined, so 5x² + 13x - 6 ≠ 0 (x + 3)(5x - 2) ≠ 0 If the product of two factors is zero, then one or both factors is zero. If x + 3 ≠ 0, x ≠ - 3 If 5x - 2 ≠ 0, 5x ≠ 2 x ≠ 2/5 Vertical Asymptotes x { - 3, 2/5 } ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ If the largest exponent in the numerator is less than the largest exponent of the denominator, the horizontal asymptote is the x-axis, so Horizontal Asymptote: y = 0 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 2.) f(x) = (x - 6) / (x + 5) Interval [- 4, - 1]: a = - 4 b = - 1 Ave. Rate of Change: _ V = [f(b) - f(a)] / (b - a) f(a) = (- 4 - 6) / (- 4 + 5) f(a) = - 10 / 1 f(a) = - 10 f(b) = (- 1 - 6) / (- 1 + 5) f(b) = - 7/4 f(b) = - 1.75 _ V = (- 1.75 -10) / [- 4 - (- 1)] _ V = - 11.75 / (- 4 + 1) _ V = - 11.75 / - 3 _ V = 3.917 ¯¯¯¯¯¯¯¯ 3.) x₁ = - 2 x₂ = 4 b = 8 a = b / (x₁)(x₂) Subbing known values, a = 8 / (- 2)(4) a = 8 / - 8 a = - 1 Equation: f(x) = - (x + 2)(x - 4) f(x) = - (x² - 2x - 8) f(x) = - x² + 2x + 8 Graph of 1 / f(x): http://s533.photobucket.com/user/revo_em...
Difference between the reciprocal of a trigonometric function & the inverse of a trigonometric function?
reciprocal - y= 1/sin(x), becomes y = cosec(x). Inverse = y = arcsin(x), you can reverse it to make it x = sin (y). Hope that explained a bit.