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wolverine [178]
3 years ago
11

[Will mark as brainliest]

Mathematics
1 answer:
Aleks04 [339]3 years ago
4 0

The answer is: f^{-1} = 4x^2-3,\quad\text{for  } x \leq 0

The inverse of a function f(x) is another function, f^{-1}(x), with the following property:

f(f^{-1}(x)) = f^{-1}(f(x)) = x

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, f(x)=2x, and one that halves it: f(x)= \frac{x}{2}. Their composition is clearly the identity function f(x)=x, since you consider "twice the half of something", or "half the double of something".

In general, to invert a function y=f(x), you have to solve the expression for x, writing an expression like x = g(y). If you manage to do so, then g is the inverse of f.

In your case, you have

f(x) = y = -\frac{1}{2}\sqrt{x+3}

Multiply both sides by -2 to get

-2y = \sqrt{x+3}

Square both sides to get

4y^2 = x+3

Finally, subtract 3 from both sides to get

x = 4y^2 - 3

Since the name of the variables doesn't really have a meaning, you can say that the inverse function is

f^{-1}(x) = 4x^2 - 3

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 b \in B such that there exists a \in A : f(a)=b. So, we need the range of f(x).

We know that the range of g(x)=\sqrt{x} is [0,\infty). When you transform it to g(x)=\sqrt{x+3} you simply translate the graph horizontally, so the range doesn't change. But when you multiply the function times -\frac{1}{2} you affect both extrema of the range, turning it into (-\infty,0], which you can simply write as x \leq 0

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