Please help ASAP ,It’s a test due in 5min
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Theoretically, the number of the vertex comes at the top must be equal for infinitely many numbers of trials, but it's a tedious job to roll the dice for a large number of times. Rolling the tetrahedron only 10 times will not predict the correct result.
Han suspects that this tetrahedron is weighted in some way to make 1 appear more often than the other numbers.
So, check his suspect experimentally with the help of the concept of center of mass. A fair tetrahedron must have the center of mass at the center of the body.
The given tetrahedron has 4 equilateral triangular faces, so at first, measure all the sides of triangular faces and make sure that all are having the same length.
Then, conduct an experiment to check the location of the center of mass.
Hang the tetrahedron by attaching a thin string at the vertex 1 as shown in the figure and observe the base surface ( triangle 234) whether it is parallel to the ground or not.
Now, repeat the same by hanging it with vertices 2, 3, and 4 and observe the base surface.
For the center of mass to be located at the center of the tetrahedron, the base must be parallel to the ground (flat ground).i.e the hanging string must be perpendicular to the base for all the four cases.
If this is so, then the tetrahedron is a fair tetrahedron otherwise not.
Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,
which is a cubic polynomial in with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).
Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.
f'(x) = 3x² - 1 = 0 ⇒ x = ±1/√3
So, we have three subsets over which f(x) can be considered invertible.
• (-∞, -1/√3)
• (-1/√3, 1/√3)
• (1/√3, ∞)
By the inverse function theorem,
where f(a) = b.
Solve f(x) = 2 for x :
x³ - x + 2 = 2
x³ - x = 0
x (x² - 1) = 0
x (x - 1) (x + 1) = 0
x = 0 or x = 1 or x = -1
Then can be one of
• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);
• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or
• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)