Solve log2 ( x - 1 ) + log2 ( x + 1 ) = 3 check answers are in domain
1 answer:
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Let's consider the functions one by one.
i)
,
x --> -infinity means that x is a very very small negative number. To model it in our mind let's think of
. This number to an even power becomes 
... etc.
Indeed, we can see that the smaller the x, the greater the value
. 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. For this value of x, y is 
, that is 
.
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, <span>f(x)-->0 as x --> -infinity
</span>
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, <span>f(x)-->0 as x --> -infinity</span>
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, <span>f(x)-->0 as x --> -infinity
</span>
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, <span>f(x)-->0 as x --> -infinity</span>