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AlekseyPX
3 years ago
6

As the domain values approach infinity, the range values approach infinity. As the domain values approach negative infinity, the

range values approach negative infinity. To which family of functions might the described function belong?
Mathematics
1 answer:
Firdavs [7]3 years ago
6 0

Limits at infinity truly are not so difficult once you've become familiarized with then, but at first, they may seem somewhat obscure. The basic premise of limits at infinity is that many functions approach a specific y-value as their independent variable becomes increasingly large or small. We're going to look at a few different functions as their independent variable approaches infinity, so start a new worksheet called 04-Limits at Infinity, then recreate the following graph.

plot(1/(x-3), x, -100, 100, randomize=False, plot_points=10001) \ .show(xmin=-10, xmax=10, ymin=-10, ymax=10) Toggle Explanation Toggle Line Numbers

In this graph, it is fairly easy to see that as x becomes increasingly large or increasingly small, the y-value of f(x) becomes very close to zero, though it never truly does equal zero. When a function's curve suggests an invisible line at a certain y-value (such as at y=0 in this graph), it is said to have a horizontal asymptote at that y-value. We can use limits to describe the behavior of the horizontal asymptote in this graph, as:

 and 

Try setting xmin as -100 and xmax as 100, and you will see that f(x) becomes very close to zero indeed when x is very large or very small. Which is what you should expect, since one divided by a large number will naturally produce a small result.

The concept of one-sided limits can be applied to the vertical asymptote in this example, since one can see that as x approaches 3 from the left, the function approaches negative infinity, and that as x approaches 3 from the right, the function approaches positive infinity, or:

 and 

Unfortunately, the behavior of functions as x approaches positive or negative infinity is not always so easy to describe. If ever you run into a case where you can't discern a function's behavior at infinity--whether a graph isn't available or isn't very clear--imagining what sort of values would be produced when ten-thousand or one-hundred thousand is substituted for x will normally give you a good indication of what the function does as x approaches infinity.

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julsineya [31]
25 + 6 - 15 + 25 - 10 + 14 - 21
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16 + 25 - 10 + 14 - 21
41 - 10 + 14 - 21
31 + 14 - 21
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3 years ago
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The regular price of an item is $150. The item is on sale at a discount rate of 30%.
Aleonysh [2.5K]

Answer:

$105

Step-by-step explanation:

30% of 150 is 45.

150-45 = 105

Hope that helps

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How do I solve 300x300=
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6 0
4 years ago
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PRECALCULUS please help​
sergejj [24]

Answer:

<h2>( f - g)(x)  =  \frac{2x -  \sqrt{x} + 14 }{3x}</h2>

Step-by-step explanation:

f(x) =  \frac{2x + 6}{3x}  \\  \\ g(x) =  \frac{ \sqrt{x}  - 8}{3x}

To find ( f - g)(x) , subtract g(x) from f(x)

That's

( f - g)(x) =  \frac{2x + 6}{3x}  -  \frac{ \sqrt{x}  - 8}{3x}

Since they have a common denominator that's 3x we can subtract them directly

That's

\frac{2x  +  6}{3x}  -  \frac{ \sqrt{x}  - 8}{3x}    =  \frac{2x  + 6 - ( \sqrt{x} - 8) }{3x}  \\  =  \frac{2x + 6 -  \sqrt{x} + 8 }{3x}  \\  =  \frac{2x -  \sqrt{x} + 6 + 8 }{3x}

We have the final answer as

<h3>( f - g)(x)  =  \frac{2x -  \sqrt{x} + 14 }{3x}</h3>

Hope this helps you

6 0
3 years ago
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