Mathematics: Post your doubts here!

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so only number 4 no need to do 1 2 3 5 6 7 8 9 and 10 thankss
just write yes if this is the case.
 
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and for drawing its graph in number 4 we just have to put the values of x lets say -4,-3,-2,-1,0,1,2,3,4 for instance put in the f(x) equation get the value of y and plot the graph is it like that thanks if not please solve if you can respont urgently please as you now i have very short time.
 
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THIS IS THE QUESTION FIND THE MAXIMUM AND THE MINMUM POINTS
F(X)=3X^4+1
 

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Assalamoalaikum!! :)

Well ...check my notes..differentiation..it's included in there..
In short, u have to differtiate the equation.... then put dy/dx = 0
you'll get the x coordinates put in the main eqn to get the value of...find d^2 y / dx^2 to find the nature...u can find more detail in my notes...as far as i remember i have this one added...try doing it yourself..it'll build your confidence too..when u understand and do and then get it correct... :) ;) if u need help...dont hesitate though ;)
 
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I HAVE SOLVED MANY PROBLEM LIKE THESE BUT THE PROBLEM IS THAT FOR D^2Y/DX^2=0 WHEN WE PUT THE VALUE OF X SO IS IT MAXIMUM OR MINIMUM OR POINT OF INFLECTION HAS A VERY SPECIAL CASE.
 

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hold on cud u put the question again... 3x^4+1

power is 4 right?
 

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Assalamoalaikum..destined007
do this type of questions come in the exam?? I never came across any such question in the pastpaper?!

thanx anyways..i never thought of this ;) jazakAllah Khairen!
 
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yeah. This can come in exam.
 

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hmm i never saw it though...but gud i know now... :)
 
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THEN WHAT ABOUT THE PROBLEM F(X)=2X^3-12X^2+24X+6 EVEN IN THIS ONE WE GET F DOUBLE DASH(X) =0 THEN WHEN WE TAKE X=-1 AND X=1 SO WE GET BOTH NEGATIVE VALUES PLEASE EXPLAIN
 
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AND SO THE ANSWER FOR THE ONE YOU POSTED IS MINIMUM AS D^2Y/DX^2 CHANGES FROM NEGATIVE TO POSITIVE MEANING MAXIMA TO MINIMUM AND THE VALUES WOULD BE (0,1)
 
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1357913579 said:
THEN WHAT ABOUT THE PROBLEM F(X)=2X^3-12X^2+24X+6 EVEN IN THIS ONE WE GET F DOUBLE DASH(X) =0 THEN WHEN WE TAKE X=-1 AND X=1 SO WE GET BOTH NEGATIVE VALUES PLEASE EXPLAIN
 

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BY THE WAY IS INFLEXION IN THE P1 SYLLABUS THANKS VERY VERY MUCH FOR ANSWERING
 
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I dont think so. However, do look at some of its questions. You never now what CIE will give in the exam.
 
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This was a question i encountered in our Co-ordinate geometry class . I found the co-ordinates of one vertex but equating the equations by there's no way except the graph work i think this can be solved !! It might seem easy to you guys but its given me a hell already !! :x

The equations of two sides of a square are y = 3x - 1 and x + 3y -6 = 0. If ( 0, -1) is one vertex of the square find the coordinates of the other vertices.

:: Core Maths 3rd Edition; Pg 91, Q 11.

Answers = ( 9/10 , 17/10 ) <-- Found it !!
Other two can be
Either ( -18/10 , 26/10 ) and ( -27/10 , -19/10 )
Or ( 36/10 , 8/10 ) and ( 27/10, -19/10) !!
 
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please help me solve this problem with all working thanks
A(7,2) and c(1,4) are two vertices of the square ABCD
(A)FIND THE EQUATION OF THE DIAGONAL BD
(B)FIND THE COORDINATES OF B AND OF D
 
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