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 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)/(cx-a) 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)/(cx-a))=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) $\displaystyle \frac{a\left(\frac{ax+b}{cx-a}\right)+b}{c\left(\frac{ax+b}{cx-a}\right) - a} = \frac{\frac{a^2x+ab}{cx-a} + b}{\frac{acx+bc}{cx-a}-a} = \frac{a^2x+ab+b(cx-a)}{acx+bc-a(cx-a)}=\frac{a^2x+bcx}{bc+a^2}=x$ 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. Thanks from Fernando89 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: 7,821 Thanks: 1047 Math Focus: Elementary mathematics and beyond a) $$(a-b)^2\ge0$$ $$a^2-2ab+b^2\ge0$$ $$\frac{a^2+b^2}{2}\ge ab$$ $$\frac{a^2+2ab+b^2}{2}\ge2ab$$ $$2\cdot\frac{a^2+2ab+b^2}{4}\ge2ab$$ $$\frac{a^2+2ab+b^2}{4}\ge ab$$ $$\frac{a+b}{2}\ge\sqrt{ab}$$
 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 AM-GM 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 $\displaystyle |a + b| = -(a + b) = -a-b=|a| + |b|$. 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 $\displaystyle |a| + |b| = -a+b \neq -a-b = |a+b|$. 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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