Describe the transformations of
f(x)=sqrtt x
that result in the functions listed below.
g(x)= -sqrt x+2
h(x)= (sqrt x-3)+4
Describe the transformations of
\(\displaystyle{f{{\left({x}\right)}}}=\sqrt{{t}}{x}\)
that result in the functions listed below.
  

\(\displaystyle{g{{\left({x}\right)}}}=-\sqrt{{x}}+{2}\)
  

\(\displaystyle{h}{\left({x}\right)}={\left(\sqrt{{x}}-{3}\right)}+{4}\)
(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).
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) solution: given that, `sinx+cosx=a`   
 on squaring both sides, we get   
 `" "(sinx+cosx)^(2)=(a)^(2)`   
 `rarr" "sin^(2)x+cos^(2)x+2sinxcosx=a^(2)`   
 `rarr" "sinx*cosx=(1)/(2)(a^(2)-1)`   
 `(i) sin^(6)x+cos^(6)x=(sin^(2)x)^(3)+(cos^(2)x)^(3)`   
 `" "=(sin^(2)x+cos^(2)x)(sin^(4)x-sin^(2)xcos^(2)x+cos^(4)x)` ltbrgt `" "=sin^(4)x+cos^(4)x-(1)/(4)(a^(2)-1)^(2)`   
 `" "=(sin^(2)x+cos^(2)x)^(2)-2sin^(2)xcos^(2)x-(1)/(4)(a^(2)-1)^(2)`   
 `" "=1-2*(1)/(4)(a^(2)-1)^(2)-(1)/(4)(a^(2)-1)^(2)=(1)/(4)[4-3(a^(2)-1)^(2)]`   
 `(ii) abs(sinx-cosx)=sqrt((sinx-cosx)^(2))`   
 `" "=sqrt(sin^(2)x+cos^(2)x-2sinxcosx)`   
 `" "=sqrt(1-2(1)/(2)(a^(2)-1))=sqrt(1-a^(2)+1)=sqrt(2-a^(2))` f(x) = u/v....i.e. u = 5x² + 4x + 3 and v = 2x² - 2x - 14

now, f '(x) = (u'v - uv')/v²

so, f '(x) = [(10x + 4)(2x² - 2x - 14) - (5x² + 4x + 3)(4x - 2)]/(2x² - 2x - 14)²   

a bit of tidying up with the denominator gives:

f '(x) = -2[9x² + 76x + 25]/(2x² - 2x - 14)²

:)> without calculus, the  horizontal coordinate of a quadratic of  f(x) =ax^2 + bx + c  is equal to - b/ 2a
   which in this case of
       f( x) =  - 5 x^2 + 3x + 2  
 
       equals :    - 3/[2(-5)] = -3/ -10 = 3/10 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) `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)` hint: mathematics includes the study of topics which are related to quantity, structure, space and change. to solve this problem, we first try to simplify the expression. for doing so we let ${{x}^{2}}-4x=t$ and solve by using this consideration. in this way we can easily factorise the expression.complete step-by-step answer:according to a question, we are given the expression $\left( {{x}^{2}}-4x \right)\left( {{x}^{2}}-4x-1 \right)-20$. to simplify this complex expression, we let ${{x}^{2}}-4x=t$.this can be mathematically expressed as: $t(t-1)-20$.since, these types of factorization are familiar to us, so we try to factorise the equation in terms of t:${{t}^{2}}-t-20$now, by using the middle term splitting method of separating the variables, we get$\begin{align}  & {{t}^{2}}+4t-5t-20=0 \\  & t(t+4)-5(t+4)=0 \\  & (t-5)(t+4)=0 \\ \end{align}$ so, the factors of the expression in terms of t are (t-5) and (t+4).now, finally replacing t as ${{x}^{2}}-4x$,we get$\left( {{x}^{2}}-4x-5 \right)\left( {{x}^{2}}-4x+4 \right)=0$ on further simplifying this expression, by using middle term splitting method again in part one:$\begin{align}  & {{x}^{2}}-4x-5={{x}^{2}}+x-5x-5 \\  & \rightarrow x(x+1)-5(x+1)=0 \\  & (x-5)(x+1)=0 \\ \end{align}$ now, for the second part, using the identity of ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, we get${{x}^{2}}-4x+4={{(x-2)}^{2}}$$\therefore (x-5)(x+1){{(x-2)}^{2}}=0$ hence, the factors of the required equation as per problem statement are: $(x-5)(x+1){{(x-2)}^{2}}=0$ note: this question can be alternatively solved by expanding the powers of x in the given expression and then letting ${{x}^{2}}$ as another variable. but the complexity of the problem would increase by doing so. hence, for simplification we must assume some other variable which can be finally replaced in the original expression to obtain the answer. solution: equation of bisectors  

`(x-2y+4)/sqrt5=(4x--3y+2)/sqrt(4^2+3^2)`  

`sqrt5x-2sqrt5y+4sqrt5=4x-3y+2`  

`y(3-2sqrt5)+x(sqrt5-4)+4sqrt5-2=0`  

`m(beta\_1)=(4-sqrt5)/(3-2sqrt5)`  

this will be negative so, this is obtuse.  

`(x-2y+4)/sqrt5=-(4x--3y+2)/sqrt(4^2+3^2)`  

in this case it will be positive so, this is acue.  
. a=c and b=d
b=d and `anec`
`ab=1 " and "cdne1`
`cd=1" and "ab=ne1`
answer: a
solution: using `a.m.gtg.m,` we have   
 `a+(1)/(b)gt2sqrt((a)/(b))b+(1)/( c )gt2sqrt((b)/( c )),c+(1)/(d)gt2sqrt(( c )/(d))" and "d+(1)/(a)gt2sqrt((d)/(a))`   
 `implies(a+(1)/(b))(b+(1)/( c ))(c+(1)/(d))(d+(1)/(a))gt16`   
 but,`(a+(1)/(b))(b+(1)/( c ))(c+(1)/(d))(d+(1)/(a))=4xx1xx4xx1=16`   
 `:." "a=(1)/(b),b=(1)/( c ),c=(1)/(d)" and "d=(1)/(a)`   
 `implies" "a=(1)/(b)=2,b=(1)/( c )=(1)/(2),c=(1)/(d)=2" and "d=(1)/(a)=(1)/(2)`   
 `implies" "a=2,b=(1)/(2),c=2" and "d=(1)/(2)`   
 `implies" "a=c" and "b=d` solution: `ch\_(4)(g) + 2o\_(2)(g) to co\_(2)(g) + 2h\_(2)o(g) + "heat. (evolved)"`