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June 27th, 2017, 12:15 PM   #1
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Smile 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
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June 27th, 2017, 12:50 PM   #2
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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 01:15 PM.
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June 27th, 2017, 01:30 PM   #3
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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
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June 27th, 2017, 02:51 PM   #4
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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'.

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June 27th, 2017, 03:16 PM   #5
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a)

$$(a-b)^2\ge0$$

$$a^2-2ab+b^2\ge0$$

$$\frac{a^2+b^2}{2}\ge ab$$

$$\frac{a^2+2ab+b^2}{2}\ge2ab$$

$$2\cdot\frac{a^2+2ab+b^2}{4}\ge2ab$$

$$\frac{a^2+2ab+b^2}{4}\ge ab$$

$$\frac{a+b}{2}\ge\sqrt{ab}$$
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June 27th, 2017, 08:11 PM   #6
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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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