# C3 differentiation question help

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#1
Hi can someone help me?

Find the turning points on the curve with equation y=(x^2)(x+1)^8
Determine the nature of each turning points and sketch the curve.

This is what I have done so far

What if the second derivative is still 0? In the answer at the back of the book, it is said this is a minimum? Why??

Thanks a lot!

Posted from TSR Mobile
0
5 years ago
#2
(Original post by pepperzealot)
Hi can someone help me?

Find the turning points on the curve with equation y=(x^2)(x+1)^8
Determine the nature of each turning points and sketch the curve.

This is what I have done so far

What if the second derivative is still 0? In the answer at the back of the book, it is said this is a minimum? Why??

Thanks a lot!

Posted from TSR Mobile
For which value of x does it say its a minimum. It only has one minimum and maximum, they occur at x=0 and x=-0.2 respectively.
0
#3
(Original post by Phichi)
For which value of x does it say its a minimum. It only has one minimum and maximum, they occur at x=0 and x=-0.2 respectively.
. But in the answer there are two minimums and one maximum.... x=-1 is also at a minimum…I
can not understand how can we know this is a minimum？ If I substitute x=-1in to the second derivative，I will get 0.
0
5 years ago
#4
(Original post by pepperzealot)
. But in the answer there are two minimums and one maximum.... x=-1 is also at a minimum…I
can not understand how can we know this is a minimum？ If I substitute x=-1in to the second derivative，I will get 0.
0
5 years ago
#5
(Original post by pepperzealot)
. But in the answer there are two minimums and one maximum.... x=-1 is also at a minimum…I
can not understand how can we know this is a minimum？ If I substitute x=-1in to the second derivative，I will get 0.
Have you tried sketching the curve by plugging in some sample values of x and seeing what happens?

It should be obvious that this function is >=0 for all values of x, so the 2 values of x where it is actually 0 (x = 0 and x = -1) must be minima. For x > 0 or x < -1 the function just gets bigger and bigger in a positive sense so the curve just increases to infinity.

The only tricky point is what happens between -1 and 0 - the function is positive in this region, so to get from 0 back down to 0 again there must be (at least) one point in between where the curve turns down again and this will be a maximum.
0
#6
(Original post by davros)
Have you tried sketching the curve by plugging in some sample values of x and seeing what happens?

It should be obvious that this function is >=0 for all values of x, so the 2 values of x where it is actually 0 (x = 0 and x = -1) must be minima. For x > 0 or x < -1 the function just gets bigger and bigger in a positive sense so the curve just increases to infinity.

The only tricky point is what happens between -1 and 0 - the function is positive in this region, so to get from 0 back down to 0 again there must be (at least) one point in between where the curve turns down again and this will be a maximum.
Thanks a lot 0
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