SOLUTION: (a) The equation of the curve is y=2x³-7x²+4x+4. Find the x value of the maximum turning point. The maximum T.P is NOT on the Y axis
So far I have that dy/dx=6x²-14x+4=0
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-> SOLUTION: (a) The equation of the curve is y=2x³-7x²+4x+4. Find the x value of the maximum turning point. The maximum T.P is NOT on the Y axis
So far I have that dy/dx=6x²-14x+4=0
div
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Question 430167: (a) The equation of the curve is y=2x³-7x²+4x+4. Find the x value of the maximum turning point. The maximum T.P is NOT on the Y axis
So far I have that dy/dx=6x²-14x+4=0
divide through by 2 3x²-7x+2=0
(3x-1)(x-2)=0
x=1/3 and x=2
When I sub 2 into y=2x³-7x²+4x+4, I get that y=0 therefore it must be x=1/3. I sub this into y=... and get 124/27.
I am unsure if this is correct.
(b) Factorise 2x³-7x²+4x+4. I used synthetic division and got that x=2 so (x-2)
∴ (x-2)(2x²-5x+2)
=(x-2)(2x-1)(x-2)
Now I am stuck. The qusetion is State the co-ordinates of the point A and hence find the values of x for which 2x³-7x²+4x+4<0.
It may be easier if I email you the question.
Any help would be greatly appreciated.
Andrew.
Hi
1) correct on (1/3,124/27)being the maximum turning point |good work
2) 2x³-7x²+4x+4 = (x-2)(2x^2-3x-2) = (x-2)(2x+1)(x-2) |x= -1/2 and x = 2
x < -1/2 lends to f(x)< 0
always recommend sketching the graph..can be extremely helpful.
can download FREE graph software at
http://www.padowan.dk.com
