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Help with C1!

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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 :redface:

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Reply 81

TSR George

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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Reply 82

B_9710

For a point to be an inflection point it is necessary that at that point

d^2y/dx^2=0

but all because

d^2y/dx^2=0

does not mean it is an inflection point necessarily.

(edited 7 years ago)

Reply 83

B_9710

Do you know what an inflection point actually is?

Reply 84

Zacken

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.

Reply 85

TSR George

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?

Reply 86

Zacken

read that back to yourself, what does it mean for the gradient to be 0?

Reply 87

Ano123

https://en.wikipedia.org/wiki/Inflection_point

Reply 88

TSR George

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 :redface:

Reply 89

TSR George

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

Reply 90

RDKGames

Study Forum Helper

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

2(x-3)^2=0

of course as the expression is always positive and anything other than 0 would put you above your minimum value.

(edited 7 years ago)

Reply 91

TSR George

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

2(x-3)^2=0

of course as the expression is always positive and anything other than 0 would put you above your minimum value.

thanks

Reply 92

TSR George

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

Reply 93

NotNotBatman

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.

Reply 94

TSR George

BINOMIAL EXPANSION help!
my text book says to use the formula

but my teacher taught me the pascal triangle which seems like less work. im just wondering if they both do the same thing, no more no less? i prefer the triangle by the looks and i really dont want to have to work with this awful long winded formula

Reply 95

TheAlphaParticle

You can't use the Triangle for everything, What if it is to the power of 8 or something like that will you draw out then. You don't have to memorise the formula learn the method and practice it often, it will become easier then.