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brinks7994
16.07.2019 •
Mathematics
Which of the following function types exhibit the end behavior f(> 0 as x --> -infinity? power; y=x^n; n is even and greater than zero identity; y=x absolute value; y= absolute value of x reciprocal; y=1/x root; y=^n sort x; n is even and greater than zero exponential; y=b^x, b> 0 again, i know that two of these are correct but i'm not sure which ones. let me know!you!
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Ответ:
i)
x --> -infinity means that x is a very very small negative number. To model it in our mind let's think of
Indeed, we can see that the smaller the x, the greater the value
ii)
y=x,
this clearly means that the behaviors of x and y are identical.
As x --> -infinity, y --> -infinity as well.
iii) y=|x|,
We can think of x --> -infinity, again, as a very very small number, like
Indeed, the smaller the x, the greater the y. As in the function in part (i), as x --> -infinity, f(x)-->+infinity.
iv)
y=1/x
consider the values of y for the following values of x:
for x=-10, -100, -1000, y is respectively -0.1, -0.01, -0.001.
We can see that the smaller the x, the closer y gets to 0.
Thus, f(x)-->0 as x --> -infinity
v) n'th root of x in not defined for negative x, when n is even.
vi) y=b^x, b>0
Here note that b cannot be equal to 1, otherwise the function is not exponential.
Let b=5, consider the values of y for the following values of x:
for x=-10, -100, -1000, y is respectively
That is, the smaller the value of x, the closer y gets to 0.
Thus, f(x)-->0 as x --> -infinity
Ответ:
I think it is in three(3) day