Rectangle B is shown below. Nadia drew a scaled version of Rectangle B using a scale factor of 1/5 and labeled it Rectangle C wh
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Given:
Area of a right triangle = 18 sq. inches
Length of one leg = 6 inches
To find:
The length of another leg.
Solution:
We know that, the area of a triangle is
In a right angle triangle, the area of the triangle is
Putting A=18 and , we get
Divide both sides by 3.
Both legs are equal so the given triangle is an isosceles right triangle.
If the triangle is not isosceles triangle, then the length of legs are factors of 36 because half of 36 is 18.
Therefore, the possible pairs of legs are 1 and 36, 2 and 18, 3 and 12, 4 and 9, 6 and 6, 9 and 4, 12 and 3, 18 and 2, 36 and 1.
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, ∞)