This is tricky for me, please help. Im still trying to learn the tangent line method of plotting new values.

Veteran  May 2, 2017

4.4.3 cannot be the derivative of any of the other functions, bc it starts negatively. For that to happen, one of the graphs would have to start out going downwards. There is no such graph, so 4.4.3 must be the answer.


5 is the derivative of 3

2 is the derivative of 5

4 is the derivative of 2

Guest May 2, 2017
edited by Guest  May 2, 2017

I think the answer is:

Figure 4.4.3 = g

Figure 4.4.5 = g'

Figure 4.4.2 = g''

Figure 4.4.4 = g'''



Look at figure 4.4.3.... the "peaks of the hills" are where the slope = 0

The peaks of the hills on this graph occur around x = 0 and x = 2.9

The slope = 0 around x = 0 and x = 2.9

The derivative = 0 around x = 0 and x = 2.9

The graph of the derivative crosses the x axis around x = 0 and x = 2.9

And figure 4.4.5 does just that! So figure 4.4.5 must be the derivative of figure 4.4.3.


Figure 4.4.5 has peaks around x = 0, x = 1.8, and x = 3.8

The derivative of this graph should cross the x-axis around x = 0, x = 1.8, and x = 3.8

Figure 4.4.2 does this, so figure 4.4.2 must be the derivative of figure 4.4.5.


Then just continue with that train of thought.. That's just how I did it! smiley

hectictar  May 2, 2017
edited by hectictar  May 2, 2017

Yes, our guest is correct.   :)  

(I though Hectictar was answering ://)


I always struggle with these myself.  I just spent quite a bit of time working it out.


2, 4 and 5 start with a positive or 0 gradient so 3 cannot be a derivative of any of them.


g    must be  4.4.3    (which I will call 3)


Now 3 has a zero, or close to zero gradient at  x=0,  and x=2.8    (they are turning points)

This is where its derivative must cross the x axis and it cannot cross any other place.

Only 5 does this.

g'   is graph 5


5 has a zero gradient at about    x=0,x=1.8 and maybe x=3.8.

So its derivative must be 2

g'' is graph 2


2 has a zero gradient at about  x=0,  x=1.2 and x=2.7

So its derivative must be  4

g''' is graph 4


It is easier to see what is happening if you line the graphs up one directly below the other.

There are other features you can use to but loking at whether the gradient is positive or negative or 0 is the main one. Looking at changes in concavity can also be used.

Melody  May 2, 2017

Arr, Hectictar was answering :)))

Melody  May 2, 2017

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