Not with the default definitions of the domain (the real numbers) and range (reals greater than or equal to -3). With these definition the original function is many-to-one (except at one point) and so the inverse function - if it existed - would be one-to-many. A function cannot be one-to-many.
However, if you redefine the domain of the original function so that it is the non-negative reals, the function becomes one-to-one and so does have an inverse which is x = +sqrt(y + 3) - the principal square root of (y+3).
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6
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One inverse is 176/r = P
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)
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Assuming that b > 0, it is an inverse power function or an inverse exponential function.
The inverse is not a function because it fails the vertical line test.
log5x
The inverse of the function y = x is denoted as y = x. The inverse function essentially swaps the roles of x and y, so the inverse of y = x is x = y. In other words, the inverse function of y = x is the function x = y.
Direct
square root of x/pi
Because the inverse of a function is what happens when you replace x with y and y with x.
No. If you invert that function, it will produce an equation that gives you two return values for one input value. This does not meet the definition of a function.
Assuming the domain and range are both the real numbers (or rationals): Yes, it is 1 to 1 Yes, it is onto and the inverse is x = (y-3)/4
5x3=15 is the inverse
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