Year 13 Maths Help Thread

Reply 220

kiiten

18

Original post by RDKGames

It is indeed 256. Also it's the area under the curve from -5 to 2, and above it from 2 to 3 (reference to the sketch).

But you can just work it out as -5 to 3 like ive done?

Original post by kiiten

But you can just work it out as -5 to 3 like ive done?

Yes that would give you the whole area enclosed by the curve and the x-axis, but your reference to the sketch is technically incorrect.

Reply 222

kiiten

18

Ughh guys ive gone wrong again :/ ques 4.b)

Posted from TSR Mobile

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Reply 223

B_9710

18

Original post by kiiten

Ughh guys ive gone wrong again :/ ques 4.b)

Posted from TSR Mobile

If the area of the 2 sections are equal, then the integral between 0 and k will be 0, as the net area above the x axis will be 0, as it will cancel with the part of the curve that is below the x axis.

Reply 224

kiiten

18

Original post by B_9710

If the area of the 2 sections are equal, then the integral between 0 and k will be 0, as the net area above the x axis will be 0, as it will cancel with the part of the curve that is below the x axis.

thanks

Original post by kiiten

thanks

Ah crap, just realised what I've said about the previous question is wrong, sorry. Anyway, k=root(8) in case you want it validated.

Reply 226

kiiten

18

Original post by RDKGames

Ah crap, just realised what I've said about the previous question is wrong, sorry. Anyway, k=root(8) in case you want it validated.

Ahh - i got the notif but never saw your post so its fine. Thanks :smile:

Original post by kiiten

Ahh - i got the notif but never saw your post so its fine. Thanks :smile:

Also your method would still work if you equate the area from 2 to k to -4 rather that positive 4, because the area is negative. :smile:

Reply 228

kiiten

18

Original post by RDKGames

Also your method would still work if you equate the area from 2 to k to -4 rather that positive 4, because the area is negative. :smile:

Eh? where did the 4 come from??

Original post by kiiten

Eh? where did the 4 come from??

That's the area from 0 to 2.

Reply 230

kiiten

18

can 4/x^-1/2 be written in a different form?

Original post by kiiten

can 4/x^-1/2 be written in a different form?

Of course.

\frac{4}{x^{-\frac{1}{2}}}=\frac{4}{\frac{1}{x^{\frac{1}{2}}}}=4x^{\frac{1}{2}}=4\sqrt{x}

(edited 7 years ago)

Reply 232

Zacken

22

Original post by kiiten

can 4/x^-1/2 be written in a different form?

\displaystyle \frac{4}{x^{-1/2}} = 4x^{1/2} = 4\sqrt{x} = \sqrt{16x} = 4 \sqrt[4]{x} \sqrt[4]{x} = \frac{4}{\frac{1}{\sqrt{x}}} = \cdots

(edited 7 years ago)

Reply 233

kiiten

18

Original post by RDKGames

Of course.

\frac{4}{x^{-\frac{1}{2}}}=\frac{4}{\frac{1}{x^{\frac{1}{2}}}}=4x^{\frac{1}{2}}=4\sqrt{x}

Original post by Zacken

\displaystyle \frac{4}{x^{-1/2}} = 4x^{1/2} = 4\sqrt{x} = \sqrt{16x} = 4 \sqrt[4]{x} \sqrt[4]{x} = \frac{4}{\frac{1}{\sqrt{x}}} = \cdots

Thanks

(sorry for asking easy ques - ive forgotten a lot of maths since the exams :3 )

Reply 234

kiiten

18

Ughhh where am i going wrong. You have to find the coordinates of the minimum. Ive completely forgotten what you have to do :frown:

(x is only 1/2)

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(edited 7 years ago)

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Reply 235

B_9710

18

Original post by kiiten

Ughhh where am i going wrong. You have to find the coordinates of the minimum. Ive completely forgotten what you have to do :frown:

(x is only -1/2 not 2 but why?)

Posted from TSR Mobile

Complete the square. Or you could differentiate and find the value of x that gives f'(x)=0, then plug in to find f at that point.

Reply 236

kiiten

18

Original post by B_9710

Complete the square. Or you could differentiate and find the value of x that gives f'(x)=0, then plug in to find f at that point.

i already found 2 values of x but they're wrong. why?

(edited 7 years ago)

Reply 237

Zacken

22

Original post by kiiten

i already found 2 values of x but they're wrong. why?

If f(x) = 6x^2 - 6x + 3 then f ' (x) = 12x - 6 so f ' (x) = 0 only when 12x - 6 = 0, not sure how you got two values of x from that.

Reply 238

kiiten

18

Original post by Zacken

If f(x) = 6x^2 - 6x + 3 then f ' (x) = 12x - 6 so f ' (x) = 0 only when 12x - 6 = 0, not sure how you got two values of x from that.

yeah i forgot to differentiate. Thanks :biggrin:

Reply 239

Zacken

22

Original post by kiiten

yeah i forgot to differentiate. Thanks :biggrin:

You even wrote "differentiation" on top! :tongue:

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