Question

Which of the following function types exhibit the end behavior f(x)-->0 as x --> -infinity?

Answer 1

Let's consider the functions one by one.

i) $y=x^n$,

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

Indeed, we can see that the smaller the x, the greater the value $x^n$. 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 $-10^{15}$. For this value of x, y is $|-10^{15}|$, that is $10^{15}$.

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 $\displaystyle{ \frac{1}{5^{10}}, \frac{1}{5^{100}}, \frac{1}{5^{1000}}$.

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

Thus, f(x)-->0 as x --> -infinity