Factor the expression: x^6-729. Show your work Please.
1 answer:
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1.) The interval of the value of x is from -5 to 1, inclusive. Remember that what is asked is the absolute value, thus the sign does not matter even if you have to subtract x from 5. Thus, the maximum value would be obtained if the x is smaller, which is 1. The minimum value is obtained when x=-5.
Absolute maximum value: x = - 5
f(-5) = ║5 - 7(-5)^2║ = ║-170║=170
Absolute minimum value: x = 1
f(1) = ║5 - 7(1)^2║ = ║-2║= 2
2.) The Mean Value Theorem (MVT) applies to functions that are continuous and differentiable on the closed and open interval of a to b, respectively. Since the function is a quadratic function, MVT can be applied. Then, this means that there is a value of c which is between a and b. This could be determined using this formula according to MVT:
The differentiated form would be f'(x) = -2x. Then,
Thus, x = -1, x = -1/2, and x=0 all lie in the function 4-x^2.
Absolute maximum value: x = - 5
f(-5) = ║5 - 7(-5)^2║ = ║-170║=170
Absolute minimum value: x = 1
f(1) = ║5 - 7(1)^2║ = ║-2║= 2
2.) The Mean Value Theorem (MVT) applies to functions that are continuous and differentiable on the closed and open interval of a to b, respectively. Since the function is a quadratic function, MVT can be applied. Then, this means that there is a value of c which is between a and b. This could be determined using this formula according to MVT:
The differentiated form would be f'(x) = -2x. Then,
Thus, x = -1, x = -1/2, and x=0 all lie in the function 4-x^2.
Check the picture below.
so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).
because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,
![\bf \begin{cases} h=2\\ k=-3\\ a=6\\ c=2\sqrt{10} \end{cases}\implies \cfrac{[y- (-3)]^2}{ 6^2}-\cfrac{(x- 2)^2}{ b^2}=1 \\\\\\ \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ b^2}=1 \\\\\\ c^2=a^2+b^2\implies (2\sqrt{10})^2=6^2+b^2\implies 2^2(\sqrt{10})^2=36+b^2 \\\\\\ 4(10)=36+b^2\implies 40=36+b^2\implies 4=b^2 \\\\\\ \sqrt{4}=b\implies 2=b\\\\ -------------------------------\\\\ \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 2^2}=1\implies \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 4}=1](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ah%3D2%5C%5C%0Ak%3D-3%5C%5C%0Aa%3D6%5C%5C%0Ac%3D2%5Csqrt%7B10%7D%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B%5By-%20%28-3%29%5D%5E2%7D%7B%206%5E2%7D-%5Ccfrac%7B%28x-%202%29%5E2%7D%7B%20b%5E2%7D%3D1%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%28y%2B3%29%5E2%7D%7B%2036%7D-%5Ccfrac%7B%28x-%202%29%5E2%7D%7B%20b%5E2%7D%3D1%0A%5C%5C%5C%5C%5C%5C%0Ac%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%282%5Csqrt%7B10%7D%29%5E2%3D6%5E2%2Bb%5E2%5Cimplies%202%5E2%28%5Csqrt%7B10%7D%29%5E2%3D36%2Bb%5E2%0A%5C%5C%5C%5C%5C%5C%0A4%2810%29%3D36%2Bb%5E2%5Cimplies%2040%3D36%2Bb%5E2%5Cimplies%204%3Db%5E2%0A%5C%5C%5C%5C%5C%5C%0A%5Csqrt%7B4%7D%3Db%5Cimplies%202%3Db%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Ccfrac%7B%28y%2B3%29%5E2%7D%7B%2036%7D-%5Ccfrac%7B%28x-%202%29%5E2%7D%7B%202%5E2%7D%3D1%5Cimplies%20%5Ccfrac%7B%28y%2B3%29%5E2%7D%7B%2036%7D-%5Ccfrac%7B%28x-%202%29%5E2%7D%7B%204%7D%3D1)
so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).
because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,