find inverse of f = 2(x - 1)^2 - 4 / find min. dimensions of

[micheleab]

1) Find the inverse for this function equation:

. . .f(x) = 2(x - 1)^2 - 4

Also how would you come up with the function for:

2) A box that has a length of the base the double of the width of the base. What dimensions should be chosen to minimize the box if the bow is to hold 1000 cubic units?

 

[soroban]

Re: functions

Hello, micheleab!

Here's the second one . . .

A box that has a length of the base the double of the width of the base.
What dimensions should be chosen to minimize the box if the bow is to hold 1000 cubic units?

I will assume that "minimize the box" means to minimize
. . the amount of cardboard used.

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            2x

Let \(\displaystyle x\) = width, then \(\displaystyle 2x\) = length.
Let \(\displaystyle y\) = height

The volume is: \(\displaystyle \,(2x)(x)(y) \,=\,1000\;\;\Rightarrow\;\;y \:=\:\frac{500}{x^2}\;\) [1]


The area of the material used to construct the box:

Top and bottom: \(\displaystyle \:2\,\times\, (2x)(x)\:=\:4x^2\)
Front and back: \(\displaystyle \:2\,\times\,(2x)(y) \:=\:4xy\)
Left and right: \(\displaystyle \:2\,\times\, (x)(y) \:=\:2xy\)

Total area: \(\displaystyle \:A\:=\:4x^2\,+\,6xy\;\) [2]


Substitute [1] into [2]: \(\displaystyle \:A \:=\:4x^2\,+\,6x\left(\frac{500}{x^2}\right)\)

. . And we have: \(\displaystyle \:A \;=\;4x^2\,+\,3000x^[-1}\)

Differentiate and equate to zero: \(\displaystyle \:8x\,-\,3000x^{-2} \;=\;0\)

Multiply by \(\displaystyle x^2:\;\;8x^3\,-\,3000\:=\:0\;\;\Rightarrow\;\;x^3\,=\,375\;\;\Rightarrow\;\;x\,=\,5\sqrt[3]{3}\)

Substitute into [1]: \(\displaystyle \:y\:=\:\frac{500}{(5\sqrt[3]{3})^2} \:=\:\frac{20\sqrt[3]{3}}{3}\)

The dimensions are: \(\displaystyle \:\left\{\begin{array}{ccc}\text{Width} & = & 5\sqrt[3]{3} \\ \\ \text{Length} & = & 10\sqrt[3]{3} \\ \\ \text{Height} & = & \frac{20\sqrt[3]{3}}{3} \end{array}\)

 

[stapel]

1) Find the inverse for this function equation:

. . .f(x) = 2(x - 1)^2 - 4

This is just algebra. Use the the methods you learned back then.

Eliz.