f(x) = 12x^3 - 117x^2 - 432x +2 List any invetervals on which the function is in
ID: 2873687 • Letter: F
Question
f(x) = 12x^3 - 117x^2 - 432x +2 List any invetervals on which the function is increasing and decreasing.
Suppose a certain drug is administered to a patient and the the percent of concentration of the drug is in the bloodstream after T hours is modeled by the following function, where 0 < T < infinity. K(t) = 4t / t^2 + 4. A) Identify the open invervals where K(t) is increasing. B) Identify the open intervals where K(t) is decreasing
The function A(x) = 0.002x^3 - 0,03x^2 +0.14x + 0.05 (The funtion only applies to the first 6 hours. A) Identify the time invervals where A(x) is increasing. B) Identify the time intervals where A(x) is decreasing
f(x) = x^3 + 6x^2 + 1. Create a sign chart to decide at which critical number a local minimum occurs and at which critical number a local maximum occurs. Find the corresponding y values for each to complete each statement.
f(x) = x^3 + 9x^2 + 24x + 1. Create a sign chart to decide at which critical number a local minimum occurs and at which critical number a local maximum occurs. Find the corresponding y values for each to complete each statement.
Explanation / Answer
1)f(x) = 12x^3 - 117x^2 - 432x +2 List any invetervals on which the function is increasing and decreasing.
f(x) = 12x^3 - 117x^2 - 432x +2
f '(x) = 12*3x^2 - 117*2x - 432+ 0
f '(x) = 36x^2 - 234x - 432+ 0
f '(x) = 2x^2 - 13x -24
f '(x) = 2x^2 - 13x -24
f '(x) = 2x^2 - 16x+3x -24
f '(x) = 2x(x-8)+3(x -8)
f '(x) = (2x+3)(x-8)
function increasing ==>f '(x)>0
==>(2x+3)(x-8)>0
==>x< -3/2 , x>8
x=(-infinity, -3/2)U(8,infinity)
function decreasing ==>f '(x)<0
==>(2x+3)(x-8)<0
==>x> -3/2 , x<8
x=( -3/2,8)
2)K(t) = 4t / t^2 + 4
K'(t) = (4*1*(t^2 +4) - 4t( 2t + 0))/ (t^2 + 4)^2
K'(t) = (4t^2 +16 -8t^2))/ (t^2 + 4)^2
K'(t) = (16 -4t^2))/ (t^2 + 4)^2
given 0<t<infinity
increasing ==>K'(t)>0
(16 -4t^2))/ (t^2 + 4)^2 >0
(16 -4t^2)) >0
(4 -t^2)) >0
(2-t)(2+t)>0
(t-2)(t+2)<0
but t>0
==>so t =(0,2)
decreasing ==>K'(t)<0
(16 -4t^2))/ (t^2 + 4)^2 <0
(16 -4t^2)) <0
(4 -t^2)) <0
(2-t)(2+t)<0
(t-2)(t+2)>0
but t>0
==>so t =(2,infinity)
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