Questions: For the following question, fill in the blank to complete the statement. (There may be more than one correct answer.)
a) If the volume of a sphere is doubled, its radius increases by a factor of .
b) If the radius of a sphere is doubled, its volume increases by a factor of .
c) The solution sets to f(x) ≤ c and x f(x) ≤ c are (the same/different).
Transcript text: For the following question, fill in the blank to complete the statement. (There may be more than one correct answer.)
a) If the volume of a sphere is doubled, its radius increases by a factor of ______.
b) If the radius of a sphere is doubled, its volume increases by a factor of ______.
c) The solution sets to f(x) ≤ c and {x | f(x) ≤ c} are (the same/different).
Solution
Solution Steps
Step 1: Solve part (a) - Volume of a sphere doubled
The volume \( V \) of a sphere is given by the formula:
\[
V = \frac{4}{3} \pi r^3
\]
If the volume is doubled, the new volume \( V' \) is:
\[
V' = 2V = 2 \left( \frac{4}{3} \pi r^3 \right) = \frac{8}{3} \pi r^3
\]
Let the new radius be \( r' \). Then:
\[
\frac{4}{3} \pi (r')^3 = \frac{8}{3} \pi r^3
\]
Divide both sides by \( \frac{4}{3} \pi \):
\[
(r')^3 = 2r^3
\]
Take the cube root of both sides:
\[
r' = \sqrt[3]{2} \, r
\]
Thus, the radius increases by a factor of \( \sqrt[3]{2} \).
Step 2: Solve part (b) - Radius of a sphere doubled
If the radius is doubled, the new radius \( r' \) is:
\[
r' = 2r
\]
The new volume \( V' \) is:
\[
V' = \frac{4}{3} \pi (r')^3 = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi (8r^3) = 8 \left( \frac{4}{3} \pi r^3 \right) = 8V
\]
Thus, the volume increases by a factor of \( 8 \).
Step 3: Solve part (c) - Solution sets comparison
The solution set to \( f(x) \leq c \) is the set of all \( x \) such that \( f(x) \leq c \). This is exactly the same as the set \( \{x \mid f(x) \leq c\} \). Therefore, the solution sets are the same.
Final Answer
a) \( \boxed{\sqrt[3]{2}} \)
b) \( \boxed{8} \)
c) The solution sets are \( \boxed{\text{the same}} \).