June 27th, 2017, 11:15 AM  #1 
Newbie Joined: Jun 2017 From: México Posts: 2 Thanks: 0 Math Focus: Algebra  Help with a demonstration
f(x)=(ax+b)/(cxa) where a, b and c are constants, with a^2+bc≠0 and c≠0 a) Find the domain b) Prove that f((ax+b)/(cxa))=x c) Find the reverse function f^1 Appreciate the help 
June 27th, 2017, 11:50 AM  #2 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics 
a) $\displaystyle \mathbb{R} \setminus \{ \frac{a}{c} \}$ b) $\displaystyle \frac{a\left(\frac{ax+b}{cxa}\right)+b}{c\left(\frac{ax+b}{cxa}\right)  a} = \frac{\frac{a^2x+ab}{cxa} + b}{\frac{acx+bc}{cxa}a} = \frac{a^2x+ab+b(cxa)}{acx+bca(cxa)}=\frac{a^2x+bcx}{bc+a^2}=x$ c) Note that a function $\displaystyle g$ is the inverse of $\displaystyle f$ if $\displaystyle (f \circ g)(x) = x$. In this case, as $\displaystyle (f \circ f)(x) = x$, $\displaystyle f$ is its own inverse. Last edited by skipjack; June 27th, 2017 at 12:15 PM. 
June 27th, 2017, 12:30 PM  #3 
Newbie Joined: Jun 2017 From: México Posts: 2 Thanks: 0 Math Focus: Algebra 
Thanks! I seize this opportunity to ask for these 2 problems I have encountered a) Prove that if a,b > 0 then √ab ≤ (a+b)/2 b) Prove that for any a,b ∈ ℝ it is fulfilled= a+b=a+b , if and only if ab≥0 
June 27th, 2017, 01:51 PM  #4 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
One way to do part a) First we show that for real numbers $ \ \ a \ \ $ , $ \ \ b \ \ $ , $ \ \ \sqrt{ab} > \frac{a + b}{2} \ \ \ \ \ $ is impossible. Then we find when $ \ > \ $ is replaced by $ \ = $ $2 \sqrt{ab} > (a + b) \ \ \ \ \ $ justified since multiplication by positive $ \ \ 2 \ \ $ does not change the direction of the inequality. $ 4ab > a^2 + 2ab + b^2 \ \ \ \ \ $ justified since $ \ \ a \ \ $ , $ \ \ b \ \ $ are positive and squaring both positive sides of an inequality does not change the direction of the inequality. $ 0 > a^2  2ab + b^2 \ \ \ \ \ $ justified since subtracting a positive real number from an inequality does not change the direction of the inequality. $ 0 > (a  b)^2 \ \ \ \ \ \ \ $ last inequality But this is impossible because the square of a real number can never be less than $ \ \ 0 \ \ $ and $ \ \ (a  b) \ \ $ is a real number. Also , we know exactly when the last inequality above is equal to $ \ \ 0 \ \ $ , when $ \ \ a = b \ \ $ and this proves the equality part of question a) Therefore it follows from the last inequality above that when $ \ \ a \ne b \ \ $ , $ \ \ 0 < (a  b)^2 $ and the entire question a) is proven. Note and further details:This approach shows that 'greater than' is impossible for question a) therefore the only possibilities remaining are 'less than or equal to'. The case 'equal to is specific , only when $ \ \ a = b \ \ $ threfore the only possibility remaining when $ \ \ a \ne b \ \ $ is 'less than'. 
June 27th, 2017, 02:16 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,570 Thanks: 931 Math Focus: Elementary mathematics and beyond 
a) $$(ab)^2\ge0$$ $$a^22ab+b^2\ge0$$ $$\frac{a^2+b^2}{2}\ge ab$$ $$\frac{a^2+2ab+b^2}{2}\ge2ab$$ $$2\cdot\frac{a^2+2ab+b^2}{4}\ge2ab$$ $$\frac{a^2+2ab+b^2}{4}\ge ab$$ $$\frac{a+b}{2}\ge\sqrt{ab}$$ 
June 27th, 2017, 07:11 PM  #6 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics 
a) also has a nifty geometric proof. My linear algebra lecturer said that he could open an entire course focused just on proving the AMGM inequality in as many ways as possible. b) This question isn't hard, but it does require you to have some experience with proof techniques. If part: Suppose $\displaystyle ab \geq 0$. Then either $\displaystyle a, b \geq 0$ or $\displaystyle a, b < 0$. If $\displaystyle a, b \geq 0$, then $\displaystyle a + b \geq 0$ and hence $\displaystyle a + b = a + b= a + b$. If $\displaystyle a, b < 0$, then $\displaystyle a + b < 0$ and hence $\displaystyle a + b = (a + b) = ab=a + b$. Only if part: Suppose $\displaystyle ab < 0$. Then $\displaystyle a$ and $\displaystyle b$ are nonzero numbers with different signs. Without loss of generality, suppose $\displaystyle a$ is negative. We now complete the proof by three cases: Case 1: Suppose $\displaystyle a > b$. Then $\displaystyle a + b < 0$, and hence $\displaystyle a + b = a b$. Yet $\displaystyle a = a$ and $\displaystyle b = b$, so $\displaystyle a + b = a+b \neq ab = a+b$. Case 2: Suppose $\displaystyle a = b$. Then $\displaystyle a + b = 0$, and hence $\displaystyle a + b = 0$. Yet $\displaystyle a = a$ and $\displaystyle b = b$, so $\displaystyle a + b = a+b \neq 0 = a+b$. Case 3: Suppose $\displaystyle a < b$. Then $\displaystyle a + b > 0$, and hence $\displaystyle a + b = a + b$.Yet $\displaystyle a = a$ and $\displaystyle b = b$, so $\displaystyle a + b = a+b \neq a+b = a+b$. This completes our proof. 

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