Finding Horizontal Pt of Inflexion (1 Viewer)

mreditor16 · May 4, 2015

Hi Guys,

Revising for a quiz soon, and got stuck on this one as well! Any help will be greatly appreciated!!

InteGrand · May 4, 2015

mreditor16 said:
Hi Guys,

Revising for a quiz soon, and got stuck on this one as well! Any help will be greatly appreciated!!

$f^\prime (x) = -4\sin(2x) -4\sin(4x-a)$

$f^{''}(x) = -8\cos(2x) -16\cos(4x-a)$

$f^{'''}(x) = 16\sin(2x) +64\sin(4x-a)$

For horizontal points of inflection, we want the following to be satisfied:

$f^\prime (x) = -4\sin(2x) -4\sin(4x-a) =0 \text{ } (1)$

$f^{''}(x) = -8\cos(2x) -16\cos(4x-a)=0 \text{ } (2)$

$f^{'''}(x) = 16\sin(2x) +64\sin(4x-a) \neq 0 \text{ } (3)$ .

Now we can see where Equations (1) and (2) take us in terms of getting suitable values of x and a, and we just need to make sure we reject any solutions that would make the third derivative 0.

InteGrand · May 4, 2015

Actually, sorry, the above that I wrote is slightly erroneous; the fully correct version of our requirements is here: http://en.wikipedia.org/wiki/Higher...general_derivative_test_for_stationary_points

InteGrand · May 4, 2015

The list of equalities does not need to stop at the second derivative, it just needs to stop at an even derivative and the next one (an odd one) must be non-zero.

integral95 · May 4, 2015

InteGrand said:
$f^\prime (x) = -4\sin(2x) -4\sin(4x-a)$

$f^{''}(x) = -8\cos(2x) -16\cos(4x-a)$

$f^{'''}(x) = 16\sin(2x) +64\sin(4x-a)$

For horizontal points of inflection, we want the following to be satisfied:

$f^\prime (x) = -4\sin(2x) -4\sin(4x-a) =0 \text{ } (1)$

$f^{''}(x) = -8\cos(2x) -16\cos(4x-a)=0 \text{ } (2)$

$f^{'''}(x) = 16\sin(2x) +64\sin(4x-a) \neq 0 \text{ } (3)$ .

Now we can see where Equations (1) and (2) take us in terms of getting suitable values of x and a, and we just need to make sure we reject any solutions that would make the third derivative 0.

He wouldn't have learnt that property yet.

He would only need equations (1) and (2) that you've mentioned

In (1) you find x in terms of a and sub into (2) and solve from there.

InteGrand · May 4, 2015

integral95 said:
He wouldn't have learnt that property yet.

He would only need equations (1) and (2) that you've mentioned

In (1) you find x in terms of a and sub into (2) and solve from there.

The problem is, Equations (1) and (2) alone wouldn't generally guarantee that it's an inflection point. We'd need to at least test our values of a and x in some way (e.g. with table of values) to show that it's not a local extremum.

Shadowdude · May 4, 2015

mreditor16 said:
Hi Guys,

Revising for a quiz soon, and got stuck on this one as well! Any help will be greatly appreciated!!

Inflextion?

wtf

photastic · May 4, 2015

Is this question any different to high school?

Finding Horizontal Pt of Inflexion (1 Viewer)

mreditor16

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InteGrand

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