find the intervals in which the following functions are increasing or decreasing.f(x) = x3 – 12x2 + 36x + 17
given:- function f(x) = x3 – 12x2 + 36x + 17theorem:- let f be a differentiable real function defined on an open interval (a,b).(i) if f’(x) > 0 for all , then f(x) is increasing on (a, b)(ii) if f’(x) < 0 for all , then f(x) is decreasing on (a, b)algorithm:-(i) obtain the function and put it equal to f(x)(ii) find f’(x)(iii) put f’(x) > 0 and solve this inequation.for the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.here we have,f(x) = x3 – 12x2 + 36x + 17⇒ ⇒ f’(x) = 3x2 – 24x + 36for f(x) lets find critical point, we must have⇒ f’(x) = 0⇒ 3x2 – 24x + 36 = 0⇒ 3(x2 – 8x + 12) = 0⇒ 3(x2 – 6x – 2x + 12) = 0⇒ x2 – 6x – 2x + 12 = 0⇒ (x – 6)(x – 2) = 0⇒ x = 2, 6clearly, f’(x) > 0 if x < 2 and x > 6and f’(x) < 0 if 2 < x < 6thus, f(x) increases on (–∞, 2) ∪ (6, ∞)and f(x) is decreasing on interval x ∈ (2, 6)
the range of the function f(x) = |x -&nbsp;1| is class intervals
frequency
(fi)
xi
di = xi - a
fi di


0 - 20
17
10
-40
-680


20 - 40
28
30
-20
-560


40 - 60
32
50 = a
0
0


60 - 80
p
70
20
20p


80 - 100
19
90
40
760


total
Î£
fi = 96 + p


Î£ fi di = 20p -
480


![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/5130830/2013_07_08_13_01_36/mathmlequation8813644107607516415%20png) a - ii, b - iv, c - i, d - iii switch x and y, then solve again for y:

x = 3y+12
x-12 = 3y
(x-12)/3 = y

or,

y = x/3 - 4 1) p(x) = 1/x2 + 1 if x ≤ 3 it is clear that x= 3 is included, so you have to use this function.2) loo k at table d, and compare the given information with the data written inside the table. ?solution 
step 1:
we have to draw the graph of form r to r. 
step 2:
the graph of f is the set of all points (x,f(x)) with x in the domain of f.
we take x in [-4,4]
and we will find the separate values of y for separate intervals.
 
 
 

 
......(1)
 

 

 

using x values along x-axis and along y-axis we get the graph of given function form r to r.
step 3:
the graph is as follow (a) g(x) is a horizontal translation of 2 units to the left then a reflection over the x-axis of f(x)
  

(b) h(x) is a horizontal translation of 3 units to the right then a vertical translation os 4 units up of graph of f(x). 31, 87, 32, 21 solution: in a given data, the highest frequency is 41, which lies in the interval 1000-15000.   
 here, l=10000, `f\_(m)=41=26, f\_(2)=16 and h=5000`   
 `"mode""=l+(f\_(m)-f\_(1))/(2f\_(m)-f\_(1)-f\_(2))xxh`   
 `=1000+((41-26)/(2xx41-26-16))xx5000`   
 `=1000+((15)/(82-42))xx5000`   
 `=1000+((15)/(40))xx5000`   
 `=1000+15xx125=10000+1875=11875`   
 hence, the modal income is 11875. find zeroes of the following polynomials (with steps ) (i) p(x)=6x^2-3-7x (ii) f(x)=2-x^2