Answer:
(3, 5) U (5, ∞)
Step-by-step explanation:
The derivative has zeros at -2, 3, and 5. It changes sign only at x=3, being zero or negative to the left of that point, and being zero or positive to the right of that point. The function increases where the derivative is positive (not zero), so on the interval ...
(3, 5) U (5, ∞)
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The function f'(x) will touch the x-axis, but not cross, at x=-2 and x=5, where the powers of the factors (x+2) and (x-5) are even. The function will cross the axis where the power of the factor is odd, at x=3.
Each place where the derivative is zero corresponds to a flat spot in the function. (The function is neither increasing nor decreasing there.) The higher the power of the corresponding factor in the derivative, the "flatter" the function is at that point.
That would be 5/8
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If there are 120 total songs and 20% are rap, then your equation is
120 * 20% = 24 rap songs
If 24 of the songs are rap and the rest are rock, then
120 - 24 = 96 rock songs
<span>Vcube = Ledge^3
8000cm^3 = Ledge^3
use the cube root of each side and you have the measure of an edge in cm.
8000^(1/3) = 20
every edge is 20 cm
20^2 = 400
The area of each face is 400 cm^2. </span>
Here is the correct format for the question
At 2:00 PM a car's speedometer reads 30 mi/h. At 2:15 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:15 the acceleration is exactly 80 mi/h².Let v(f) be the velocity of the car t hours after 2:00 PM.Then 
. By the Mean Value Theorem, there is a number c such that 0 < c < 
with v'(c) = . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly 80 mi/h^2.
Answer:
Step-by-step explanation:
From the information given :
At 2:00 PM ;
a car's speedometer v(0) = 30 mi/h
At 2:15 PM;
a car's speedometer v(1/4) = 50 mi/h
Given that:
v(f) should be the velocity of the car t hours after 2:00 PM
Then will be:
= 20 × 4/1
= 80 mi/h²
By the Mean value theorem; there is a number c such that :
with 
