hectictar

Thank you! There are many different ways to explain the problem with dividing by zero, but I like that way. Maybe someone else will provide another explanation.

Here's one way to figure out, for example, what is  10 ÷ 2

We had to subtract  2  a total of  5  times before we couldn't subtract it any more (without getting a number < 0)

Or we can say that  2  "fits into"  10  a total of  5 times.

So  10 ÷ 2  =  5

Now let's try to use the same method to figure out what is  10 ÷ 0  

We keep subtracting  0  over and over again, but we aren't getting anywhere!  This is an endless loop.

There is not a number of times that we can subtract  0  from  10  to reach a number < 0.

In other words, there is not a number that is equal to  10 ÷ 0

Yes, that is correct!

Here's a graph of the equation   h  =  -16t² + 64t + 31

where you can see that the maximum value of  h  is  95  and it occurs when  t = 2

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:

h  =  -16t² + 64t + 31

maximum value of  h  will occur when    t  =  -b / [2a]

that is, when   t  =  -64 / [ 2(-16) ]

Once you know the value of  t  that produces the maximum value of  h ,

plug that value in for  t  to find the height at that time.

Can you finish it? If you get stuck let us know.

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.

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

Assuming the question is "What is the maximum value of  c  such that. . ."

\(y\ =\ \frac13x^2\\~\\ y\ =\ x+c\)

Equate both expressions of  y

\(\frac13x^2\ =\ x+c\\~\\ \frac13x^2-x-c\ =\ 0\)

This quadratic equation will have only one solution when the discriminant equals zero.

\((-1)^2-4(\frac13)(-c)\ =\ 0\\~\\ 1+\frac43c\ =\ 0\\~\\ \frac43c\ =\ -1\\~\\ c\ =\ -\frac34\)

It should be  4 pi 6^2    then later    14400/36  =  400

And can you explain why do you get the LCM of those three numbers?