Question

If u(x)=-2x²+3 and v(x)=1/x, what is the range of (u ° v)(x)?

Answer 1

$u(x)=-2x^2+3\\\\v(x)=\dfrac{1}{x}\\\\(u\ \circ\ v)(x)=-2\left(\dfrac{1}{x}\right)^2+3=-2\left(\dfrac{1}{x^2}\right)+3=-\dfrac{1}{x^2}+3\\\\\text{The range of}\ y=\dfrac{1}{x^2}\ \text{is all positive real numbers.}\\\\\text{The range of}\ y=-\dfrac{1}{x^2}\ \text{is all negative real numbers.}\\\\\text{The range of }\ (u\ \circ\ v)(x)=-\dfrac{1}{x^2}+3\ is\ (-\infty,\ 3)$

Answer 2

Answer:

(-∞,3)

Step-by-step explanation:

Imagine functions as little machines that turn one number into another number, the set of numbers that can enter the machine are called domain and the set of  numbers that can exit the machine are called range.  In this excercise we have to do function composition, which would be like having two machines and take the numbers that exit machine number 2 ($v(x)$) and enter them into the machine number 1 ($u(x)$). In math this is done by replacing the $x$ in $u(x)$ with the function $v(x)$ like this:

$u(x)=-2x^{2} +3\\v(x)=\frac{1}{x}\\(u°v)=-2(\frac{2}{x})^{2} +3$

Now we need the range of the composed function, this will be the numbers that can come out of the machine number 1 when the numbers from the machine number 2 are entered to it.

So first which numbers will come out of machine number 2($v(x)$)? All but 0 because  we there is no number that we can divide 1 by it that will give us the value 0 (not even zero itself because it is and indeterminate form).

We have now which numbers will enter machine number 1 (we dont have any restrictions in $u(x)$ to enter numbers)

The range of the composed function will be then the range of $u(x)$ less the value that we woud obtain by replacing $x$ with 0.

The range of $u(x)$ is (-∞,3] according to the  attached graph and the value that we woud obtain by replacing $x$ with 0 is 3 so we would have (-∞,3).