Question

If u (x) = negative 2 x squared and v (x) = StartFraction 1 Over x EndFraction, what is the range of (u circle v) (x)?
(one-third, 0)
(3, infinity)
(negative infinity, 3)
(negative infinity, positive infinity)

Answer 1

Answer:

The range is negative numbers.

The interval for the range is $(-\infty,0)$.

***You might want to look at your functions again because I don't see a choice that matches.

Step-by-step explanation:

Given functions:

$u(x)=-2x^2$

$v(x)=\frac{1}{x}$

We are asked to find the range of $(u \circ v)(x)$.

I'm also going to look at the domain just to see if this possibly might change my range .

$v(x)$ is the inner function. So we will consider the domain of that function first.

You only have to worry about division by zero for the function $v$.

Since we are dividing by $x$, we don't want $x$ to be zero.

So far the domain is all real numbers except $x=0$.

Now let's move out.

$u(x)=-2x^2$ exists for all numbers, $x$. So we didn't want to include $x=0$ from before.

Now let's put it together:

$(u \circ v)(x)$

$u(v(x))$

$u(\frac{1}{x})$

$-2(\frac{1}{x})^2$

$-2(\frac{1^2}{x^2})$

$-2(\frac{1}{x^2})$

$\frac{-2}{x^2}$

So the domain is still all real numbers except at $x=0$ since we cannot divide by 0 and $x^2$ is 0 when $x=0$.

$y=\frac{-2}{x^2}$ with $x \neq 0$.

$x^2$ is positive for all numbers except $x=0$.

So $\frac{-2}{x^2}$ is negative for all numbers since negative divided by positive is negative.

So the range is only negative numbers.

Let's also look at the inverse:

$y=\frac{-2}{x^2}$

Multiply both sides by $x^2$:

$yx^2=-2$

Divide both sides by $y$:

$x^2=\frac{-2}{y}$

Take the square root of both sides:

$x=\pm \sqrt{\frac{-2}{y}}$.

So $y$ can't be 0 and it also can't be positive because the inside of the square root will be negative (since negative divided by positive results in negative).

Answer 2

Answer:

C.) (negative infinity, 3)

Step-by-step explanation:

Plz trust this is correct