If u(x)=-2x^2+3 and v(x)=1/x what is the range of (u*v)(x) **the picture should help

Mathematics

2 answers:

nasty-shy [4]3 years ago

5 0

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

using a graph tool
see the attached figure

The horizontal asymptote of this function is at y=3.
So,
the range of this function is from (−∞,3)

il63 [147K]3 years ago

5 0

Answer:

-OO,3

Step-by-step explanation:

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Find derivative problem Find B’(6)

dalvyx [7]

Answer:

$B^\prime(6) \approx -28.17$

Step-by-step explanation:

We have:

$\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)$

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

$\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]$

We can move the constant outside:

$\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]$

Now, we will utilize the product rule. The product rule is:

$(uv)^\prime=u^\prime v+u v^\prime$

We will let:

$\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t$

Then:

$\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1$

(The derivative of u was determined using the chain rule.)

Then it follows that:

$\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}$