The correct structure of the question is as follows:
The function f(x) = x^3 describes a cube's volume, f(x) in cubic inches, whose length, width, and height each measures x inches. If x is changing, find the (instantaneous) rate of change of the volume with respect to x at the moment when x = 3 inches.
Answer:
Step-by-step explanation:
Given that:
f(x) = x^3
Then;
V = x^3
The rate whereby V is changing with respect to time is can be determined by taking the differentiation of V
dV/dx = 3x^2
Now, at the moment when x = 3;
dV/dx = 3(3)^2
dV/dx = 3(9)
dV/dx = 27 cubic inch per inch
Suppose it is at the moment when x = 9
Then;
dV/dx = 3(9)^2
dV/dx = 3(81)
dV/dx = 243 cubic inch per inch
Answer:
Step-by-step explanation:
7+3-2=3+7-2
Answer:
$4.26
Step-by-step explanation:
63.90 divide by 15 = $4.26
the price of one tube is $4.26
Answer:
f(x) = 0.43 * 
* 
*(x + 10)
Step-by-step explanation:
We have a 6th degree polynomial f(x)
r = 3 is a root of f with multiplicity 2
r = 1 is a root of f with multiplicity 3
f(-5) = -29721.6
f(-10) = 0
Then: f(x) = a*((x -3)^2 ) * ((x - 1)^3)*(x + 10)
f(-5) = a * (-8)^2 * (-6)^3 * (5) = -29,721.6
a* (64) * (-216)* 5 = -29,721.6
-a*69,120 = -29,721.6
a = -29,721.6/-69,120
a = 0.43
so
f(x) = 0.43 * * *(x + 10)
