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`V(x) = (350x)/(1+x^2)^(3/4)`
We're trying to find effectively what V(x) approaches as x becomes very small. The best way to do this would be to think of what happens as x becomes very very small.
The numerator would still be `350x`. There is no problem there. However, the denominator starts to look more and more like it's approaching 1 as x gets closer and closer to zero!
To find out why, let's look at it closer:
Our denominator is `(1+x^2)^(3/4)`
We have to look at what number is dominating this expression. Because we know that x << 1, we know that `x^2` <<<< 1! For example, if `x = 0.01` then `x^2 = 0.0001` and the denominator is:
This looks pretty much like `1^(3/4) = 1`.
Another example: x = 0.00001. Well, then `x^2 = 0.0000000001` and `(1+x^2)^3/4 = 1.0000000001^(3/4)`
We'll, again, just call this 1.
Do you see what I mean? As x gets closer and closer to 0, the 1 is the main factor in the denominator. We can repeat these examples for any number of x's close to zero, and we get the same result: something that looks a heck of a lot like 1.
So, we'll just say that if x is much less than 1, our denominator goes from `(1+x^2)^(3/4)` to `1^(3/4)`. However, 1 to the 3/4 power is just 1. Therefore, our denominator is pretty much equal to 1!
Another look, as x approaches 0, we change the function on the left to the function on the right:
`(350x)/(1+x^2)^(3/4) -> (350x)/(1^(3/4)) = 350x`
And that is why, when x gets much smaller than 1, V(x) starts to look like 350x.
I hope that works!
We want to show that V(x) is approximated by f(x) = 350x near x = 0.
If this were the case, then V'(0) = 350. Let's see if that's true.
Looks like a mess. We could clean it up...or just plug in 0 to find `V'(0)=350(1)-0=350`
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