Question 1210389
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Answer: <font color=red>true</font>


Explanation


I think you meant to type
f(x) = 1/x


As x gets bigger, i.e. as we move to the right, what happens to y = 1/x?


We can look at a graph to note how it goes downhill throughout its entire domain. 
Therefore the answer is <font color=red>true</font>
{{{graph(400,400,-5,5,-5,5,-100,1/x)}}}
As x gets bigger, y gets smaller.
As we move to the right the curve goes downhill.


A way to do this without a graph is to plug in a value like x = 2
1/x = 1/2 = 0.50
Then plug in x = 3
1/x = 1/3 = 0.33 approximately
Comparing 0.50 with 0.33 shows that y has gone down, which helps show the function decreases. 
Keep in mind that this does not prove it always decreases over the entire domain. 
You'll have to use the method mentioned in the next paragraph.


Let's do a formal proof.
Let a > 0 and b > 0.
Since both are positive this means a+b > 0.
Plug in x = a to arrive at 1/a which is also positive.
Plug in x = a+b to arrive at 1/(a+b) which is also positive.
Now to compare 1/a with 1/(a+b)
Assume for now they are equal,
1/a = 1/(a+b)
a+b = a .... cross multiply
b = 0
We arrive at a contradiction since we made b > 0 earlier.
To fix those errors, replace each equal sign with a greater than sign
1/a > 1/(a+b)
a+b > a .... cross multiply
b > 0
Or basically we move through those steps backwards like so
b > 0
a+b > a
1/a > 1/(a+b)
This basically shows that for some input x = a, if we nudge that input to the right a bit to arrive at x = a+b, then we have shown that 1/a is larger than 1/(a+b)
In short we have proven the function is <u>always</u> going downhill no matter where you're on it.
Note that we can follow very similar steps when a < 0 and b > 0, to effectively mirror things over the y axis. 
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