the range of the function f(x) = |x &#8211;&nbsp;1| is &#8230;.
the range of the function f(x) = |x -&nbsp;1| is
the range of the function f(x) = |x -&nbsp;1| is
?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 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) Let $y=f(x)$ be a function. Suppose that $a$ and $\mathbf{L}$ are numbers such that:


 


* whenever $\boldsymbol{x}$ is close to $\boldsymbol{a}$ but not equal to $\boldsymbol{a}, \boldsymbol{f}(\boldsymbol{x})$ is close to $\mathbf{L} ;$
* as $\boldsymbol{x}$ gets closer and closer to $\boldsymbol{a}$ but not equal to $\boldsymbol{a}, \boldsymbol{f}(\boldsymbol{x})$ gets closer and closer to $\mathbf{L}$; and
* suppose that $f(x)$ can be made as close as we want to $\mathbf{L}$ by making $x$ close to $a$ but not equal to $a$.


Then we say that the limit of $f(x)$ as $x$ approaches a is $L$ and we write $\lim \_{x \rightarrow a} f(x)=\mathbf{L}$ doing integration yields-then finally f(x)=-x cos x + sin xto calculate derivative of the above function f(x) we have to apply formula of derivation of products of two functions-f(x) =-x cos(x)+sin(x)f’(x) =-[x(-sin x)+cos x.1]+cos x ;{by formula d/dx (f.g)=f.g’+g.f’}f’(x) =-(-x sin x) - cos x + cos xf’(x) = x.sin(x) 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) formula used/theory.there is some value of ‘x’ for which the f(x) comes to 0, that value of ‘x’ is said to be zero of polynomial f(x)we have,f(x) = 2 – xthen,zero of the polynomial f(x)means f(x) = 02 – x = 0-x = -2x = 2∴ zero of polynomial f(x) is 2conclusion.hence, the zero of the polynomial f(x) is 2 the relationship of wavelength and frequency will appear as:using the wavelength range given, 280-320 nm, frequency range will be calculated using the equation above and by converting the values to the proper units [readmore]each wavelength must be in meters:we need to calculate the frequency for both 280 and 320 nm:range of frequency will appear as 1.07x1015 to 9.38x1014 s-1 `f(x)=log(logx),g (x)=(1)/(logx)`
`f(x)=logx,g(x)=(1)/(logx)`
`f(x)=(1)/(logx),(x)=log(logx)`
`f(x)=(1)/(xlogx),g(x)=(1)/(logx)`
answer: a
solution: we have , `i=int{log(logx)+(1)/((logx)^(2))}dx`   
 `rarr i= inte^(t)(logt+(1)/(t^(2)))dt`,where t = log x   
 `rarri=inte^(t)(logt+(1)/(t))dt+inte^(t)(-(1)/(t)+(1)/(t^(2)))dt`   
 `rarri=e^(t)logt +e^(t)(-(1)/(t))+c`   
 `rarri=x(log(logx)-(1)/(logx))+c`   
 `:. f(x) =logx and g(x) = (1)/(logx)` (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). y = 3x

It can be observed that x = −1, y = −3 and x = 1, y = 3 are solutions of the above equation. Therefore, the solution table is as follows.



| x | -1 | 1 |
| y | -3 | 3 |

The graph of the above equation is constructed as follows.

![](https://www.sarthaks.com/?qa=blob&qa_blobid=15681428413536225923)