+0  
 
0
46
13
avatar+748 

Hi friends, last question for the day..maybe for a while..wink..

 

\(f(x)=3x^3+bx^2+cx-27\)

 

\(f'(1)=12\)   -   I presume this indicates one of the turning points, to be (1;12)

\(f''(1)=-24\)   -   i presume this to mean the point of inflection is at (1;-24)

 

Calculate b and c.

 

Thats it!, to all my friends, Melody, Alan and ElectricPavlov..thank you for helping me out over the last day or 2....I wish I could return the favours!!.smiley

 Oct 14, 2020
 #1
avatar+12483 
+1

Calculate b and c.

ElectricPawlow's answer is right.

laugh

 
 Oct 14, 2020
edited by Omi67  Oct 15, 2020
 #5
avatar+748 
+1

Hi Omi67,

I see your answers are different to ElectricPavlov's?...Thanx for your time..

 
juriemagic  Oct 14, 2020
 #8
avatar+27695 
+1

Omi made a small math mistake here:

   - 9        + c   = 36    |: ( - 9)      

 

should be a + 9 resulting in c = 45   

    then the b calculation will change to b = -21

 
ElectricPavlov  Oct 14, 2020
 #10
avatar+748 
0

Thank you ElectricPavlov...smiley

 
juriemagic  Oct 15, 2020
 #2
avatar+27695 
+2

f' = first derivative         = 9x^2 + 2bx + c

f" = second derivative  = 18x +2b

 

f" (1) =- 24                18(1)+ 2b = -24     then b = -21

 

Use this valueof b in the first equation

f'(1) = 12 = 9(1^2) + 2(-21)(1) + c = 12    

                    9    +  -42     + c = 12       c = 45

 
 Oct 14, 2020
 #7
avatar+748 
0

ElectricPavlov,

 

thank you very much!..much appreciated!

 
juriemagic  Oct 14, 2020
 #3
avatar+111124 
+2

No that is not what it means at all.

 

It just means that when x=1 the gradient of the tangent will be 12

 

And when x=1 the concavity will be -24.   Only the negative sign needs to mean anything to you.

the concavity at x=1 is negative so at that point will be on a  concave down part of the graph.

 

I can see that EP is composing an answer as well.

He will probably show you the algebra of what you have to do.

I will come back later if he does not fill in the holes to your satisfaction.  Let us know how you get on.  :)

 

 

 

 

 

 
 Oct 14, 2020
 #4
avatar+111124 
+1

ALWAYS REMEMBER:

 

The first derivative at any given point gives the GRADIENT of the tangent to the curve at that point.

 

It is essential that you understand and memorize this fact.

 
 Oct 14, 2020
 #6
avatar+748 
+1

Melody,

 

yes, I understand..brrr...okay, it's a lot right now, I wil have a look at all the replies a bit later on...my brain is smoking!..blush..but..I will check it and I hereby also say a BIIIG thank you!..

 
juriemagic  Oct 14, 2020
 #9
avatar+748 
+1

Melody..good morning,

Just to mention, the reason I looked at the 1st derivative as being one of the turning points, is that we do use the 1st derivitave also to calculate the x-coordinates of the turning points, by, once the derivitave is calculated, it is then factored to get the x-coordinates, and it was because of that, that I thought they gave us one of those turning points. but I will see it from now that they mean to supply us with gradient information....Thank you..smiley

 
juriemagic  Oct 15, 2020
 #11
avatar+111124 
+1

Yes, At the turning points the gradient of the tangent must be 0

so

the first derivative at that point must be zero.

 

this is because

 

THE GRADIENT OF THE TANGENT IS GIVEN BY THE FIST DERIVATIVE

 
Melody  Oct 15, 2020
 #12
avatar+748 
+1

oh yes...off course..You know Melody....I am working through some grade 12 papers and a lot of things are starting to really open up and sink in..I simply cannot believe that things are not as difficult as they seem, it really is common sense in sooo many cases, but having said this, I really want to thank ALL of you guys who have remained patient throughout my silly questions, and in many cases having explained things more than once. I finally understand all of this now..laugh

 
juriemagic  Oct 15, 2020
 #13
avatar+111124 
0

That is great Juriemagic. cool

 
Melody  Oct 15, 2020

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