What is this please answer
1 answer:
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If you wanted to rotate f(x) around the y axis from x=a to x=b then the volume would be
that would be
1/x^10=x^-10
integrate
-1/(9x^9)
so
=
![\pi [\frac{-1}{9x^{9}}]^8_1=\pi (\frac{-1}{9(8^{9})}-\frac{-1}{9(1^{9})})=](https://tex.z-dn.net/?f=%5Cpi%20%5B%5Cfrac%7B-1%7D%7B9x%5E%7B9%7D%7D%5D%5E8_1%3D%5Cpi%20%28%5Cfrac%7B-1%7D%7B9%288%5E%7B9%7D%29%7D-%5Cfrac%7B-1%7D%7B9%281%5E%7B9%7D%29%7D%29%3D)
that's the volume of the solid
that would be
1/x^10=x^-10
integrate
-1/(9x^9)
so
that's the volume of the solid
Answer:
(-√8, -√6] ∪ [-√2, 0) ∪ (0, √2] ∪ [√6, √8)
Step-by-step explanation:
The inequality resolves into 4 inequalities. There are 4 intervals in the solution.
Starting at the left, for the absolute value argument less than 0:
2 ≤ -(x^2 -4) . . . . . . . for x^2 -4 ≤ 0
2 ≤ -x^2 +4
-2 ≤ -x^2
2 ≥ x^2 . . . . . . . . . . consistent with the above 4 ≥ x^2
-√2 ≤ x ≤ √2 . . . . . square root; may be limited by other constraints
For the absolute value argument greater than 0:
2 ≤ x^2 -4 . . . . . . . for x^2 -4 ≥ 0
6 ≤ x^2 . . . . . . . . . .consistent with x^2 ≥ 4
-√6 ≥ x ∪ x ≤ √6 . . . . take the square root
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The inequality on the right can be written as the compound inequality ...
-4 < x^2 -4 < 4
0 < x^2 < 8 . . . . . add 4
0 < |x| < √8 . . . . take the square root
This resolves to ...
-√8 < x < 0 ∪ 0 < x < √8
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So, the solution set is the set of values of x that satisfy these restrictions on x:
-√2 ≤ x ≤ √2
x ≤ -√6 ∪ x ≤ √6
-√8 < x < 0 ∪ 0 < x < √8
That is a collection of 4 intervals:
(-√8, -√6] ∪ [-√2, 0) ∪ (0, √2] ∪ [√6, √8)
_____
You may be expected to write √8 as 2√2.
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These intervals are the portions of the red curve that lie between the two horizontal lines. The points on the upper (dashed) line are not part of the solution set. The points on the lower (solid) line are part of the solution set.
