No.
An exponential function can be is of the form f(x) = a*(b^x). Some examples are f1(x) = 3*(10^x), or f2(x) = e^(-2*x). Note that the latter still fits the format, with b = e^(-2). The inverse is the logarithmic function. So for y = f1(x) = 3*(10^x), reverse the x & y, and solve for y:x = 3*(10^y)log(x) = log(3*(10^y)) = log(3) + log(10^y) = log(3) + y*log(10) = y*1 + log(3)y = log(x) - log(3) = log(x/3)The second function: y = e^(-2*x), the inverse is: x = e^(-2*y).ln(x) = ln(e^(-2*y)) = -2*y*ln(e) = -2*y*1y = -ln(x)/2 = ln(x^(-1/2))See related link for an example graph.
First, this function is strictly increasing on the entire real line, so an inverse exist on the entire real line. We define inverse of function f, denoted f^-1 such that if y = f(x) then f^-1(y) = x Or to find the inverse, all is needed is to isolate x in terms of y. In this case, y = 7x + 2 7x = y - 2 x = (y - 2)/7 So the inverse is x = (y - 2)/7 What? You don't like function in terms of y? Well, they are just meaningless variables anyway, you can write whatever, in particular the inverse is y = (x - 2) / 7 (the x, y here are independent with the x, y above. If you are getting confused, write b = (a - 2)/7 where b is a function of a)
Both. If you look at it like this: y=41-x you say x is independent and y is dependent of x (i.e. y is a function of x) For x=41-y you say y is independent and x is dependent of y (i.e. x is a function of y)
y0(x) could represent a function of x but usually y(0) represents the function y that is evaluated at x = 0 and so is no longer a function of x but a constant.
X - Y^2 = 1 - Y^2 = - X + 1 Y^2 = X - 1 Y = (+/-) sqrt(X - 1) now, X is represented as a function of Y. Function values are generally Y values.
f ( x ) = (x-2)/(x-1)if y = (x-2)/(x-1)yx-y= x - 2yx-x= -2+yx(y-1)=y-2x = (y-2)/(y-1)so g ( x ) the inverse function is also (x-2)/(x-1)
A function of x means that the answers depends on the value of x that is substituted into the function. For example: y = x + 1 If x = 1, y = 2 If x = 2, y = 3 and so on.
Yes, y=-2+x is a linear function.
Yes it does, Remember Y values are generally function values. So, putting a value into this function, substitution a integer for X, fives you the Y value. Y = X + 4 ( make X 2 ) Y = (2) + 4 Y = So, when X = 2, Y = 6. The function.
x + y = -2 y = -x - 2 f(x)= - x -2
no, but y = x2 is a function
Simply stated, the inverse of a function is a function where the variables are reversed. If you have a function f(x) = y, the inverse is denoted as f-1(y) = x. Examples: y=x+3 Inverse is x=y+3, or y=x-3 y=2x+5 Inverse is x=2y+5, or y=(x-5)/2
The "root" of a function is also called the "zero" of a function. This is where the function equals zero. The function y=4-x2 has roots at x=2 and x=-2 The function y=4-x2 has zeroes at x=2 and x=-2 Those are equivalent statements.
y = -x-2
No.
The inverse of a function (G(x)) can be found by switching the roles of (x) and (y) and solving for (y). Given the function (G(x) = -\frac{4}{3}x + 2), let's find its inverse: Step 1: Replace (G(x)) with (y): [y = -\frac{4}{3}x + 2] Step 2: Swap (x) and (y): [x = -\frac{4}{3}y + 2] Step 3: Solve for (y): [x - 2 = -\frac{4}{3}y] [-\frac{3}{4}(x - 2) = y] So, the inverse function (G^{-1}(x)) is: [G^{-1}(x) = -\frac{3}{4}(x - 2)]