Derivatives

[TWELVEPEANUTS11]

The question asks to arrange the following numbers in increasing order and explain your reasoning.

o, g'(-2), g' (0), g'(2), g'(4)

there is a graph that has intersection points on x axis at -2, 0 and at 2. and goes from (-infinity, infinity)

the answer says g' (0), 0, g'(4), g'(2), g'(-2)

I am not exactly sure how they came up with that exact order, so if anyone could assist me that would be excellent.

 

[mmm4444bot]

Obviously, any helper will need to see the graph. Your description of the exercise is inadequate.

I mean, the graph is a graph of what ?

 

[soroban]

Hello, TWELVEPEANUTS11!

Arrange the following numbers in increasing order and explain your reasoning.
. . \(\displaystyle 0,\;g'(-2),\; g' (0),\; g'(2),\; g'(4)\)

There is a graph that has intersection points on \(\displaystyle x\)-axis at \(\displaystyle -2,\: 0,\:2\), and goes from (-\infty,\:\infty)[/tex]

The answer says: .\(\displaystyle g'(0),\;0,\;g'(4),\;g'(2),\; g'(-2)\)

I am not exactly sure how they came up with that exact order.
Without seeing the graph, I'm not exactly sure either.

I can take a guess at what the graph looks like . . .

                  |
                  |                             *
           *      |                     o
        *      *  |                *    :
      *          *|            *        :
  - -o- - - - - - o - - - - -o- - - - - + - - - - - - 
    -2            | *     *  2          4
    *             |    *
                  |

\(\displaystyle \text{The slope at }x = 0\text{ is negative, say, }g'(0) = -1\)

\(\displaystyle \text{The number 0 (zero) is next.}\)

\(\displaystyle \text{The slope at }x = 4\text{ is slightly positive, say, }g'(4) = \tfrac{1}{2}\)

\(\displaystyle \text{The slope at }x = 2\text{ is positive and steeper, say, }g'(2) = 1\)

\(\displaystyle \text{The slope at }x = -2\text{ is even steeper, say, }g'(-2) = 2\)