A shipping container is in the shape of a right rectangular prism with a length of 10
1 answer:
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Solve for <em>x</em> when √(<em>x</em> ² - 4) = 1 :
√(<em>x</em> ² - 4) = 1
<em>x</em> ² - 4 = 1
<em>x</em> ² = 5
<em>x</em> = ±√5
We're looking at <em>x </em>≤ 0, so we take the negative square root, <em>x</em> = -√5.
This means <em>f</em> (-√5) = 1, or in terms of the inverse of <em>f</em>, we have <em>f</em> ⁻¹(1) = -√5.
Now apply the inverse function theorem:
If <em>f(a)</em> = <em>b</em>, then (<em>f</em> ⁻¹)'(<em>b</em>) = 1 / <em>f '(a)</em>.
We have
<em>f(x)</em> = √(<em>x</em> ² - 4) → <em>f '(x)</em> = <em>x</em> / √(<em>x</em> ² - 4)
So if <em>a</em> = -√5 and <em>b</em> = 1, we get
(<em>f</em> ⁻¹)'(1) = 1 / <em>f '</em> (-√5)
(<em>f</em> ⁻¹)'(1) = √((-√5)² - 4) / (-√5) = -1/√5
The sign must be negative; see the attached plot, and take note of the negatively-sloped tangent line to the inverse of <em>f</em> at <em>x</em> = 1.
Fill your inequality in with the y and x provided and then do the math. (4, -1) would fill in like this (I will use brackets to indicate absolute value symbols, since there are none in the equation editor): ![-1\ \textgreater \ [4]-5](https://tex.z-dn.net/?f=-1%5C%20%5Ctextgreater%20%5C%20%5B4%5D-5)
The right side is in fact equal to the left side so that's not the answer. For (-1, -4): ![-4\ \textgreater \ [-1]-5](https://tex.z-dn.net/?f=-4%5C%20%5Ctextgreater%20%5C%20%5B-1%5D-5)
and these are also equal. Let's try C now (-4, 1): ![1\ \textgreater \ [-4]-5](https://tex.z-dn.net/?f=1%5C%20%5Ctextgreater%20%5C%20%5B-4%5D-5)
. The absolute vale of -4 is 4 so 4 - 5 = -1 which is, in fact, less than 1. So C is our answer.