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hectictar

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Usernamehectictar
Score8718
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Questions 8
Answers 2764

 #4
avatar+8718 
+3

For a function to be invertible, it must pass the "horizontal line test," which means you must not be able to draw a horizontal line that intersects the graph more than once. We can see the graph of  y = f(x)  by itself does not pass the horizontal line test and so is not invertible.

 

 

Let's imagine what  g(x)  would look like if  a = 1

g(x)  =  f(x) + ax       If   a = 1   then

g(x)  =  f(x) + x

 

Let's see what  g(1)  would be.

g(1)  =  f(1) + 1         And from the picture we can see   f(1)  =  -5

g(1)  =  -5 + 1

g(1)  =  -4                  So the point  (1, -4)  is on the graph of  g(x)

Adding the x-value to the y-value ended up shifting that point up.

 

Let's see what  g(3)  would be.

g(3)  =  f(3) + 3         And from the picture we can see  f(3)  =  2

g(3)  =  2 + 3

g(3)  =  5                  So the point  (3, 5)  is on the graph of  g(x)

Adding the x-value to the y-value ended up shifting that point up  (even more than last time).

 

Let's see what  g(-1)  would be.

g(-1)  =  f(-1) - 1           And from the picture we can see  f(-1)  =  3

g(-1)  =  3 - 1

g(-1)  =  2                     So the point  (-1, 2)  is on the graph of  g(x)

Adding the x-value to the y-value ended up shifting the point down.

 

Let's see what  g(-2)  would be.

g(-2)  =  f(-2) - 2           And from the picture we can see  f(-2)  =  5

g(-2)  =  5 - 2

g(-2)  =  3                     So the point  (-2, 3)  is on the graph of  g(x)

Adding the x-value to the y-value ended up shifting that point down (even more than last time).

 

And if we increase the value of  a , that will only exaggerate the changes.

 

By now we can see what  g(x)  should look like compared to  f(x)

Here is a graph of  f(x)  and  g(x)  with a slider for different values of  a:

 

https://www.desmos.com/calculator/eoqgwwn620

 

Moving the slider to the right until the graph "straightens out" enough to pass the horizontal line test, we can see that the smallest positive integer that makes  g(x)  invertible is  5. Moving the slider to the left (and zooming out as necessary), we can see that the largest negative integer that makes g(x)  invertible is  -4.

 

(-4)2 + 52   =   16 + 25   =   41

Jul 31, 2019
 #2
avatar+8718 
+3

Also...

 

since the value must be the same for all nonzero real numbers  a  and  b  such that  |a| ≠ |b|

 

then we can choose for instance  a = 1  and  b = 2  and evaluate the expression.

 

\(\phantom{=\quad}\left( \frac{b^2}{a^2} + \frac{a^2}{b^2} - 2 \right) \times \left( \frac{a + b}{b - a} + \frac{b - a}{a + b} \right) \times \left( \frac{\frac{1}{a^2} + \frac{1}{b^2}}{\frac{1}{b^2} - \frac{1}{a^2}} - \frac{\frac{1}{b^2} - \frac{1}{a^2}}{\frac{1}{a^2} + \frac{1}{b^2}} \right) \\~\\ {=\quad}\left( \frac{2^2}{1^2} + \frac{1^2}{2^2} - 2 \right) \times \left( \frac{1 + 2}{2 - 1} + \frac{2 - 1}{1 + 2} \right) \times \left( \frac{\frac{1}{1^2} + \frac{1}{2^2}}{\frac{1}{2^2} - \frac{1}{1^2}} - \frac{\frac{1}{2^2} - \frac{1}{1^2}}{\frac{1}{1^2} + \frac{1}{2^2}} \right)\\~\\ {=\quad}\left(\frac41+\frac14-2 \right) \times \left( \frac{3}{1} + \frac{1}{3} \right) \times \left( \frac{\frac11 + \frac{1}{4}}{\frac{1}{4} - \frac{1}{1}} - \frac{\frac{1}{4} - \frac{1}{1}}{\frac{1}{1} + \frac{1}{4}} \right)\\~\\ {=\quad}\left(\frac94 \right) \times \left( \frac{10}{3} \right) \times \left( \frac{\frac54 }{ - \frac34} - \frac{-\frac34}{\frac54} \right)\\~\\ {=\quad}\left(\frac94 \right) \times \left( \frac{10}{3} \right) \times \left( \frac{5 }{ - 3} - \frac{-3}{5} \right)\\~\\ {=\quad}\left(\frac94 \right) \times \left( \frac{10}{3} \right) \times \left( -\frac{16}{15}\right)\\~\\ {=\quad}\left(\frac11 \right) \times \left( \frac{2}{1} \right) \times \left( -\frac{4}{1}\right)\\~\\ {=\quad}-8\)

 

 

And to make triple-sure, here's what WolframAlpha's resultsmiley

Jul 29, 2019