The answer is:
The inverse of a function is another function,
, with the following property:
In other words, the inverse of a function does exactly "the opposite" of what the original function does, and so if you compute them both in sequence you return to the starting point.
Think for example to a function that doubles the input, , and one that halves it:
. Their composition is clearly the identity function
, since you consider "twice the half of something", or "half the double of something".
In general, to invert a function , you have to solve the expression for
, writing an expression like
. If you manage to do so, then
is the inverse of
.
In your case, you have
Multiply both sides by to get
Square both sides to get
Finally, subtract 3 from both sides to get
Since the name of the variables doesn't really have a meaning, you can say that the inverse function is
As for the domain of the inverse function, remember what we said ad the beginning: if the original function goes from set A (domain) to set B (codomain), then the inverse function goes from set B (domain) to set A (codomain). This means that the inverse function is defined on an element in B if and only if that element belongs to the range of the original function, i.e. the set of the elements of the codomain such that there exists
. So, we need the range of
.
We know that the range of is
. When you transform it to
you simply translate the graph horizontally, so the range doesn't change. But when you multiply the function times
you affect both extrema of the range, turning it into
, which you can simply write as
Substitute the value n = 4 in the expressions2nn + 23(n+4)
Answers:
- Part a)
- Part b) 3
- Part c) 2.24
- Part d) 1.58
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Work Shown:
Part (a)
The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)
The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3
We apply the distance formula to say...
We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.
Saying is the same as writing d = sqrt(x^2-(x^2-3)^2)
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Part (b)
Plug in x = 0 and you should find the following
This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).
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Part (c)
Repeat for x = 1
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Part (d)
Graph the d(x) function found back in part (a)
Use the minimum function on your graphing calculator to find the lowest point such that x > 0.
See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.
The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)
This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.
