MATH SOLVE

9 months ago

Q:
# which of the following function types exhibit the end behavior f(x)-->0 as x --> -infinity?power; y=x^n;n is even and greater than zeroidentity; y=xabsolute value; y= absolute value of x reciprocal;y=1/xroot; y=^n sort x; n is even and greater than zeroexponential; y=b^x, b>0again, I know that two of these are correct but I'm not sure which ones. Please let me know! Thank you!

Accepted Solution

A:

Let's consider the functions one by one.

i) [tex]y=x^n[/tex],

x --> -infinity means that x is a very very small negative number. To model it in our mind let's think of [tex]-10^{15}[/tex]. This number to an even power becomes [tex]10^{30}, 10^{60}[/tex]... etc.

Indeed, we can see that the smaller the x, the greater the value [tex]x^n[/tex]. In fact, as x --> -infinity, f(x)-->+infinity.

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 [tex]-10^{15}[/tex]. For this value of x, y is [tex]|-10^{15}|[/tex], that is [tex]10^{15}[/tex].

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 [tex] \displaystyle{ \frac{1}{5^{10}}, \frac{1}{5^{100}}, \frac{1}{5^{1000}}[/tex].

That is, the smaller the value of x, the closer y gets to 0.

Thus, f(x)-->0 as x --> -infinity

i) [tex]y=x^n[/tex],

x --> -infinity means that x is a very very small negative number. To model it in our mind let's think of [tex]-10^{15}[/tex]. This number to an even power becomes [tex]10^{30}, 10^{60}[/tex]... etc.

Indeed, we can see that the smaller the x, the greater the value [tex]x^n[/tex]. In fact, as x --> -infinity, f(x)-->+infinity.

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 [tex]-10^{15}[/tex]. For this value of x, y is [tex]|-10^{15}|[/tex], that is [tex]10^{15}[/tex].

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 [tex] \displaystyle{ \frac{1}{5^{10}}, \frac{1}{5^{100}}, \frac{1}{5^{1000}}[/tex].

That is, the smaller the value of x, the closer y gets to 0.

Thus, f(x)-->0 as x --> -infinity