hectictar

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Usernamehectictar
Score8842
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Questions 8
Answers 2806

 #1
avatar+8842 
+3

If     f( -x )  =  f(x)     then the function is even.

If     f( -x )  =  -f(x)    then the function is odd.

 

\(f(x) = \dfrac{5^x - 1}{5^x + 1}\)

                                       Plug in  -x  for  x

\(f(-x) = \dfrac{5^{(-x)} - 1}{5^{(-x)} + 1}\)

                                       Now we are looking for a way rewrite the right side so that  f(x)  appears.

                                       Let's rewrite  5(-x)  as  1 / 5x

\(f(-x) = \dfrac{\frac{1}{5^x} - 1}{\frac{1}{5^x} + 1}\)

                                             Multiply the numerator and denominator by  5x

\(f(-x) = \dfrac{\frac{1}{5^x} - 1}{\frac{1}{5^x} + 1}\cdot\dfrac{5^x}{5^x}\)

                                             Distribute the  5x  to the terms in the numerator and denominator

\(f(-x) = \dfrac{1 - 5^x}{1 + 5^x}\)

                                             Factor  -1  out of the numerator.

\(f(-x) = \dfrac{-1(-1 + 5^x)}{1 + 5^x}\)

                                             Addition can be done in any order so we can rearrange the terms like this..

\(f(-x) = \dfrac{-1( 5^x-1)}{ 5^x+1}\)

                                             Now we can write the  -1  beside the fraction like this...

\(f(-x) = -1\cdot\dfrac{ 5^x-1}{ 5^x+1}\)

                                             Finally, f(x)  has appeared!  \(f(x) = \dfrac{5^x - 1}{5^x + 1}\)   so we can substitute  f(x)  in for  \(\dfrac{5^x - 1}{5^x + 1}\)

\(f(-x) = -1\cdot f(x)\)

 

\(f(-x) = - f(x)\)

 

Remember that if     f( -x )  =  -f(x)    then the function is odd.

 

Since  f( -x )  =  -f(x)  , the function is odd.

Jul 27, 2019
 #2
avatar+8842 
+4

As Rom said, there is definitely no way to put 70 pictures on the board.

 

68 pictures is the maximum possible, and here is a way to do that:

 

 

However, if you can't rotate the pictures 90 degrees, then the maximum possible is

 

floor( 32 / 3 )  *  floor( 32 / 5 )   =   10 * 6   =   60

Jul 27, 2019
 #1
avatar+8842 
+3

x2 + 5x  <  6

                            Subtract  6  from both sides of the inequality.

x2 + 5x - 6  <  0

                            Let's find what values of  x  make  x2 + 5x - 6  equal  0

x2 + 5x - 6  =  0

                            Factor the left side. What two numbers add to  5  and multiply to  -6 ?   -1  and  +6

(x - 1)(x + 6) = 0

                            Set each factor equal to zero and solve for  x

x - 1  =  0     or     x + 6  =  0

 

 

x  =  1                   x  =  -6

 

   

Since a graph of   y  =  x2 + 5x - 6  is a parabola, we can be sure that

 

 

the values of  x  that would make  y < 0  fall in one of these two intervals:

 

either   the interval  (-6, 1)   or   the interval  (-∞, -6) U (1, ∞)

 

 

   

To determine which interval is the solution set, let's test a number in both of them.

 

   
0  is a number in the interval  (-6, 1)

 

 

If   x  =  0   then   y  =  (0)2 + 5(0) - 6  =  -6

 

And   -6 < 0   so we know  0  should be included.

 

 

   
2  is a number in the interval  (-∞, -6) U (1, ∞)

 

 

If   x  =  2   then   y  =  (2)2 + 5(2) - 6  =  8

 

And  2  > 0   so we know  2  should not be included.

 

 

   
So we can be sure that   x2 + 5x - 6  <  0   if and only if  x  is in the interval  (-6, 1)  

 

Check: https://www.desmos.com/calculator/ovccl7zguc

Jul 26, 2019