That's easy. Take any function that is defined on R except 0, then shift it 4.1 units to the right. To do a shift operation of a units, it is simply taking f(x - a) instead of f(x)
So in this case, take f(x) = 1/x, a is 4.1
f(x - 4.1) = 1/ (x - 4.1) is exactly what you want.
Also, there are other piecewise functions:
f(x) = 1 if x < 4.1 and 0 if x > 4.1 not defined when x = 4.1
It could be either depending on the function that you have.
The domain of a function encompasses all of the possible inputs of that function. On a Cartesian graph, this would be the x axis. For example, the function y = 2x has a domain of all values of x. The function y = x/2x has a domain of all values except zero, because 2 times zero is zero, which makes the function unsolvable.
The domain of the function f(x) = (x + 2)^-1 is whatever you choose it to be, except that the point x = -2 must be excluded. If the domain comes up to, or straddles the point x = -2 then that is the equation of the vertical asymptote. However, if you choose to define the domain as x > 0 (in R), then there is no vertical asymptote.
No. The modulus function maps two values (except 0) from the domain (-x, and x) to one value (+x) in the range or codomain. This means that for the inverse mapping each value in the new domain (the original codomain) is associated with two values in the new codomain (original domain). A function cannot map one value to more than one.
That all depends on the functions you are given for the problem! When you add or subtract from the original function, then we obtain the new function. If you combine the functions with different domains, then the domain of the addition/subtract obeys whatever domain other function has. For instance: f(x) = 1/x and g(x) = x f(x) + g(x) = (1/x) + x = (1 + x²)/x Then, the domain is all real values except 0 since x = 0 makes the denominator zero, hence making the whole expression undefined. If x = 0, then this makes f(0) undefined.
The domain of y = 1/x2 is all numbers from -infinity to + infinity except zero. The range is all positive numbers from zero to +infinity, except +infinity.
The cotangent function has domain all real numbers except integral multiples of pi./2(90degrees).
The domain of a rational function is the whole of the real numbers except those points where the denominator of the rational function, simplified if possible, is zero.
All real numbers except 2
It could be either depending on the function that you have.
The domain is all real numbers except when the denominator equals zero: x2 - 4 = 0 x2 = 4 x = 2, -2 So the domain is all real numbers except 2 and -2.
Let the function be f(x) = 1/(x-1) The domain is all allowable values for which the function can be defined. Here, except 1, any number would give the function a meaningful value. If x=1, the denominator becomes 0 and the function becomes undefined. Therefore, the domain is all real numbers except 1. The range is all values assumed by the function. Here, the range is negative infinity to plus infinity (that is , all real numbers).
Domain = set of points at which a function exists Range = set of points which are mapped to by the function For example, if f(t) = 1/t => domain is all real numbers except for t=0, since 1/0 is an mathematical error and: range is all real numbers except for f(t) = 0 since you cannot actually obtain this value by inputting a value for t.
f(x)=(x/|x-3|)+1; domain is all real numbers except 3. f(x)=(x/(|x-3|+1)); domain is all real numbers.
The domain is all real numbers except 0.
4
yes, bcoz evey function gives some output for input. Except constant function.