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Archimedes solved this in his famous book <em>Sphere and Cylinder</em> in 225 BC. The result is reportedly on his tombstone.
The ratio of the volume of the radius r sphere to the cylinder radius r, height 2r is
The same ratio is the ratio of the surface area of the sphere to that of the cylinder (including the ends).
Answer: Last choice
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a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)
b. The inverse image of g(x) is
c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
<h3 /><h3>The domain of T</h3>Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0
⇒ x² ≥ 25
⇒ x ≥ ±5
⇒ -5 ≤ x ≤ 5.
<h3>Inverse image of f(x)</h3>
The inverse image of f(x) is f⁻¹(x) = ∛(x/3)
f : R → R defined by f(x) = 3x³
Let f(x) = y.
So, y = 3x³
Dividing through by 3, we have
y/3 = x³
Taking cube root of both sides, we have
x = ∛(y/3)
Replacing y with x we have
y = ∛(x/3)
Replacing y with f⁻¹(x), we have
So, the inverse image of f(x) is f⁻¹(x) = ∛(x/3)
<h3>Inverse image of g(x)</h3>
The inverse image of g(x) is
g : R+ → R defined by g(x) = ln(x).
Let g(x) = y
y = ln(x)
Taking exponents of both sides, we have
Replacing x with y, we have
Replacing y with g⁻¹(x), we have
So, the inverse image of g(x) is
The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
h : R → R defined by h(x) = x − 9
Let y = h(x)
y = x - 9
Adding 9 to both sides, we have
y + 9 = x
Replacing x with y, we have
x + 9 = y
Replacing y with h⁻¹(x), we have
So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
Learn more about inverse image of a function here: