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1. (Original post by Zacken)
wot?

for a stationary point to be an inflection point, you need f'(x) = 0 and f''(x) = 0.

[and the obvious caveat for anybody who cares]
This is what I dont get

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2. (Original post by RDKGames)
I'll have some of whatever drugs you're having, mate.

Stationary points where f'(x)=0
Points of inflection where f''(x)=0

Combine the two for both to be true.

"dy/dx must be positive/negative and NOT change its sign (-/+)" ...huh?
"so what i do is the 'gradient table'..." ...why? Just solve for values of x then solve for values of x on f''(x)=0. Compare the two for the same x's and you've found the x coordinates of the points which are stationary AND points of inflection at the same time.
I dont know tbh what is the example in my book on about

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3. (Original post by ihatePE)
Hello, i just want a confirmation relating to ''determining whether a stationary point is a point of inflection''

when f''(x) = 0 you need to investigate further right?
so what i do is the 'gradient table' where i find out what dy/dx is when x=-1 x=0 x=1 ect

my point is, for it to be a point of inflection, dy/dx must be positive/negative and NOT change its sign (-/+) when x=-1 and x=1... am i right?

i've read the notes in the book for ages and its worded so briefly that im not so sure if that's the rules.
For a point to be an inflection point it is necessary that at that point but all because does not mean it is an inflection point necessarily.
4. (Original post by ihatePE)
I dont know tbh what is the example in my book on about

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Do you know what an inflection point actually is?
5. In particular, to back up B_9710's comment, you'll see that you need the third derivative to be non-zero. if you're not convinced, look at x^4.
6. (Original post by B_9710)
Do you know what an inflection point actually is?
it's when the gradient on either side of the stationary point is 0?
7. (Original post by ihatePE)
it's when the gradient on either side of the stationary point is 0?
read that back to yourself, what does it mean for the gradient to be 0?
8. (Original post by ihatePE)
it's when the gradient on either side of the stationary point is 0?
https://en.wikipedia.org/wiki/Inflection_point
9. (Original post by Zacken)
read that back to yourself, what does it mean for the gradient to be 0?
oh stationary... inflection is when it doesnt change on either side of the stationary point so like positive on left and positive on right/ neg on left and right? something like that ??
10. complete the square question

a)express x^2 + 6x - 4 in the form (x+a)^2 + b
b) use your results to part a) to find the least value of 2x^2 + 12x - 8 and the corresponding value of x

i know how to do part a), but im not sure what part b) is even asking. i talked about this with my teacher, she said take the 2 out as the correspondant
2 [ x^2 + 6x - 4 ]
but i still dont get what she want me to do afterwards
11. (Original post by ihatePE)
complete the square question

a)express x^2 + 6x - 4 in the form (x+a)^2 + b
b) use your results to part a) to find the least value of 2x^2 + 12x - 8 and the corresponding value of x

i know how to do part a), but im not sure what part b) is even asking. i talked about this with my teacher, she said take the 2 out as the correspondant
2 [ x^2 + 6x - 4 ]
but i still dont get what she want me to do afterwards
Okay, so you factored out the 2. Now put the completed square form inside the bracket. Expand (the outer bracket) and find when the value is a minimum. That will be when of course as the expression is always positive and anything other than 0 would put you above your minimum value.
12. (Original post by RDKGames)
Okay, so you factored out the 2. Now put the completed square form inside the bracket. Expand (the outer bracket) and find when the value is a minimum. That will be when of course as the expression is always positive and anything other than 0 would put you above your minimum value.
thanks
13. show that
x^2 + (2k - 1)x + (k^2 - k + 2) = 0
has no real roots, whatever the value of the constant k (4 marks)

and find the range of values of x satisfying the inequality
3x^2 + 16x - 12 > 0

for the first part i got to

''for no real roots b^2 - 4ac < 0
4k^2 + 1 - 4(k^2-k+2)
-4k+9=0
k=-9/-4
hence k<0

is that it? im confused at the wording of the question
and the second quetion is part b) and usually when this happen, it links with the first part of the question, i just want to know if there's a link? if not then i dont need help with the second question, i can do that
14. (Original post by ihatePE)
show that
x^2 + (2k - 1)x + (k^2 - k + 2) = 0
has no real roots, whatever the value of the constant k (4 marks)

and find the range of values of x satisfying the inequality
3x^2 + 16x - 12 > 0

for the first part i got to

''for no real roots b^2 - 4ac < 0
4k^2 + 1 - 4(k^2-k+2)
-4k+9=0
k=-9/-4
hence k<0

is that it? im confused at the wording of the question
and the second quetion is part b) and usually when this happen, it links with the first part of the question, i just want to know if there's a link? if not then i dont need help with the second question, i can do that
You squared the bracket wrong and be careful with the signs, for there to be no real roots the discriminant has to be negative; there's no k terms in the discriminant.

For the second part, factorise and find the critical values.

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