What is BD?

A square with side of 10 has area of 100. a. True b. False

iris [78.8K]

The answer is True tell me if i am right.

4 0

2 years ago

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[Will mark as brainliest]

Aleks04 [339]

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 $x \leq 0$