Please Help me in my maths problems (1 Viewer)

jadenmaccas · Jul 23, 2011

Hey I'm in yr 10 doing accelerated math and I want to know the definition of the point of inflection in calculus. Our teacher went ahead and we were going through it today but I missed a lesson recently and was confused. I need to ask some questions:

(1) What is the point of inflection?
(2) How do I find the point of inflection?
(3) Say I had a curve: 2x^{3}+4x^{2}+6x-17
May you show me the steps to sketch this curve by ONLY finding maximum and minimum, and calculus and the y-intercept, not finding the x-intercepts?

Thank you very much!

Alkanes · Jul 23, 2011

1) Change in concavity
2) At inflexion pt, y'' = 0 (Differentiate the equation twice)
3) Sorry cbf right now going sleep

kthx bai

Shadowdude · Jul 23, 2011

jadenmaccas said:
Hey I'm in yr 10 doing accelerated math and I want to know the definition of the point of inflection in calculus. Our teacher went ahead and we were going through it today but I missed a lesson recently and was confused. I need to ask some questions:

(1) What is the point of inflection?
(2) How do I find the point of inflection?
(3) Say I had a curve: 2x^{3}+4x^{2}+6x-17
May you show me the steps to sketch this curve by ONLY finding maximum and minimum, and calculus and the y-intercept, not finding the x-intercepts?

Thank you very much!

3. You find the maximum and minimum points (I assume you know how to do that) and that is the range of the function. Your polynomial is defined for all real x so your function isn't limited to a particular domain - so you might want to take limits to infinity to see where it goes.

The y-intercept is where the function hits the y-axis obviously so solve your curve for y = 0.

After that, it's just connecting the dots.

SpiralFlex · Jul 23, 2011

Example 1: Lets find the maximum and minimum points.

$y=2x^3+4x^2+2x-5$

There are two ways you can do this with Calculus!

Way number one:

First of all take the derivative of our function.

$\frac{dy}{dx}=6x^2+8x+2$

$=3x^2+4x+1$

For maximum and minimum turning points,

$\frac{dy}{dx}=0$

$3x^2+4x+1=0$

$(3x+1)(x+1)=0$

$x= -\frac{1}{3}$ , $-1$

Substitute this into the original equation,

$\Rightarrow y=2(-\frac{1}{3})^3+4(-\frac{1}{3})^2+2(-\frac{1}{3})-5$

$y=\frac{127}{27}$

$\Rightarrow y=2(-1)^3+4(-1)^2+2(-1)-5$

$y=5$

So our points are

$(-\frac{1}{3}, \frac{127}{27})$ and $(-1, 5)$

Now, to test their nature (max or min), we can take a test point very close to that $x$ coordinate on the left or right side.

Consider $x=-\frac{1}{3}$

Now what is a number significantly close to $-\frac{1}{3}$ on the left side?

Let's say, $-0.4$

So we substitute this into,

$(3x+1)(x+1)=0$

Hence, $(3(-0.4)+1)((-0.4)+1)$

We get a negative answer!

Now let's substitute a point to the right!

Let's say $-0.2$ ,

So,

$(3(-0.2)+1)((-0.2)+1)$

We get a positive answer!

So, change from negative to positive is a minimum turning point!

Hence,

$(-\frac{1}{3}, \frac{127}{27})$ is a minimum turning point.

Let's now do,

At $x=-1$

So taking a number left of it, $-1.1$

Substitute it back in,

$(3(-1.1)+1)((-1.1)+1)$

We get a positive number!

Now let's find a number to the right, $-0.9$

$(3(-0.9)+1)((-0.9)+1)$

We get a negative!

Change from positive to negative is a maximum turning point!

Way number two:

$\frac{dy}{dx}=3x^2+4x+1$

For maximum and minimum,

$\frac{dy}{dx}=0$

$3x^2+4x+1=0$

We already did this!

$(-\frac{1}{3}, \frac{127}{27})$ and $(-1, 5)$

Take the second derivative,

$\frac{d^2y}{dx^2}=6x+4$

So now we substitute the $x$ values inside.

$\frac{d^2y}{dx^2}=6(-\frac{1}{3})+4 > 0$

If $\frac{d^2y}{dx^2} > 0$ a minimum turning point occurs.

So it is a minimum turning point.

Now the other $x$ value,

$\frac{d^2y}{dx^2}=6(-1)+4 <0$

If $\frac{d^2y}{dx^2} < 0$ a maximum turning point occurs.

So it is a maximum turning point.

Now for POSSIBLE points of inflexion,

$\frac{d^2y}{dx^2}=0$

$6x+4=0$

$x=-\frac{2}{3}$

Substitute it into our original equation,

We get a POSSIBLE inflexion at $(-\frac{2}{3}, -\frac{139}{27})$

So we need to test this,

Consider,

$x=-\frac{2}{3}$

Substitute a point significantly close to the left and right,

Let's do to the left, $-0.7$

$\frac{d^2y}{dx^2}=6(-0.7)+4$

Negative!

Now a number to the right, $-0.6$

$\frac{d^2y}{dx^2}=6(-0.6)+4$

Positive!

Now a change of concavity has occurred, so we can say that,

$(-\frac{2}{3}, -\frac{139}{27})$ is a point of inflexion.

powlmao · Jul 23, 2011

SpiralFlex said:
Maximum/minimum:

$y=2x^3+4x^2+2x-5$

There are two ways you can do this with Calculus!

Way number one.

First of all take the derivative of our function.

$\frac{dy}{dx}=6x^2+8x+2$

$=3x^2+4x+1$

For maximum and minimum turning points,

$\frac{dy}{dx}=0$

$3x^2+4x+1=0$

$(3x+1)(x+1)=0$

$x= -\frac{1}{3}$ , $-1$

Substitute this into the original equation,

$\Rightarrow y=2(-\frac{1}{3})^3+4(-\frac{1}{3})^2+2(-\frac{1}{3})-5$

$y=\frac{127}{27}$

$\Rightarrow y=2(-1)^3+4(-1)^2+2(-1)-5$

$y=5$

So our points are

$(-\frac{1}{3}, \frac{127}{27})$ and $(-1, 5)$

Now, to test their nature (max or min), we can take a test point very close to that $x$ coordinate on the left or right side.

Consider $x=-\frac{1}{3}$

Now what is a number significantly close to $-\frac{1}{3}$ on the left side?

Let's say, $-0.4$

So we substitute this into,

$(3x+1)(x+1)=0$

Hence, $(3(-0.4)+1)((-0.4)+1)$

We get a negative answer!

Now let's substitute a point to the right!

Let's say $-0.2$ ,

So,

$(3(-0.2)+1)((-0.2)+1)$

We get a positive answer!

So, change from negative to positive is a minimum turning point!

Hence,

$(-\frac{1}{3}, \frac{127}{27})$ is a minimum turning point.

Let's now do,

At $x=-1$

So taking a number left of it, $-1.1$

Substitute it back in,

$(3(-1.1)+1)((-1.1)+1)$

We get a positive number!

Now let's find a number to the right, $-0.9$

$(3(-0.9)+1)((-0.9)+1)$

We get a negative!

Change from positive to negative is a maximum turning point!

Second way:

$\frac{dy}{dx}=3x^2+4x+1$

For maximum and minimum,

$\frac{dy}{dx}=0$

$3x^2+4x+1=0$

We already did this!

$(-\frac{1}{3}, \frac{127}{27})$ and $(-1, 5)$

Take the second derivative,

$\frac{d^2y}{dx^2}6x+4$

So now we substitute the $x$ values inside.

$\frac{d^2y}{dx^2}6(-\frac{1}{3})+4 > 0$

So it is a minimum turning point,

Now the other $x$ value,

$\frac{d^2y}{dx^2}6(-1)+4 <0$

So it is a maximum turning point.

Now for POSSIBLE points of inflexion,

$\frac{d^2y}{dx^2}=0$

$6x+4=0$

$x=-\frac{2}{3}$

Substitute it into our original equation,

We get a POSSIBLE inflexion at $(-\frac{2}{3}, -\frac{139}{27})$

So we need to test this,

Consider,

$x=-\frac{2}{3}$

Substitute a point significantly close to the left and right,

Let's do to the left, $-0.7$

$\frac{d^2y}{dx^2}=6(-0.7)+4$

Negative!

Now a number to the right, $-0.6$

$\frac{d^2y}{dx^2}=6(-0.6)+4$

Positive!

Now a change of concavity has occurred, so we can say that,

$(-\frac{2}{3}, -\frac{139}{27})$ is a point of inflexion.

Fucking beast!!!

Hermes1 · Jul 23, 2011

SpiralFlex said:
Maximum/minimum:

$y=2x^3+4x^2+2x-5$

There are two ways you can do this with Calculus!

Way number one.

First of all take the derivative of our function.

$\frac{dy}{dx}=6x^2+8x+2$

$=3x^2+4x+1$

For maximum and minimum turning points,

$\frac{dy}{dx}=0$

$3x^2+4x+1=0$

$(3x+1)(x+1)=0$

$x= -\frac{1}{3}$ , $-1$

Substitute this into the original equation,

$\Rightarrow y=2(-\frac{1}{3})^3+4(-\frac{1}{3})^2+2(-\frac{1}{3})-5$

$y=\frac{127}{27}$

$\Rightarrow y=2(-1)^3+4(-1)^2+2(-1)-5$

$y=5$

So our points are

$(-\frac{1}{3}, \frac{127}{27})$ and $(-1, 5)$

Now, to test their nature (max or min), we can take a test point very close to that $x$ coordinate on the left or right side.

Consider $x=-\frac{1}{3}$

Now what is a number significantly close to $-\frac{1}{3}$ on the left side?

Let's say, $-0.4$

So we substitute this into,

$(3x+1)(x+1)=0$

Hence, $(3(-0.4)+1)((-0.4)+1)$

We get a negative answer!

Now let's substitute a point to the right!

Let's say $-0.2$ ,

So,

$(3(-0.2)+1)((-0.2)+1)$

We get a positive answer!

So, change from negative to positive is a minimum turning point!

Hence,

$(-\frac{1}{3}, \frac{127}{27})$ is a minimum turning point.

Let's now do,

At $x=-1$

So taking a number left of it, $-1.1$

Substitute it back in,

$(3(-1.1)+1)((-1.1)+1)$

We get a positive number!

Now let's find a number to the right, $-0.9$

$(3(-0.9)+1)((-0.9)+1)$

We get a negative!

Change from positive to negative is a maximum turning point!

Second way:

$\frac{dy}{dx}=3x^2+4x+1$

For maximum and minimum,

$\frac{dy}{dx}=0$

$3x^2+4x+1=0$

We already did this!

$(-\frac{1}{3}, \frac{127}{27})$ and $(-1, 5)$

Take the second derivative,

$\frac{d^2y}{dx^2}6x+4$

So now we substitute the $x$ values inside.

$\frac{d^2y}{dx^2}6(-\frac{1}{3})+4 > 0$

So it is a minimum turning point,

Now the other $x$ value,

$\frac{d^2y}{dx^2}6(-1)+4 <0$

So it is a maximum turning point.

Now for POSSIBLE points of inflexion,

$\frac{d^2y}{dx^2}=0$

$6x+4=0$

$x=-\frac{2}{3}$

Substitute it into our original equation,

We get a POSSIBLE inflexion at $(-\frac{2}{3}, -\frac{139}{27})$

So we need to test this,

Consider,

$x=-\frac{2}{3}$

Substitute a point significantly close to the left and right,

Let's do to the left, $-0.7$

$\frac{d^2y}{dx^2}=6(-0.7)+4$

Negative!

Now a number to the right, $-0.6$

$\frac{d^2y}{dx^2}=6(-0.6)+4$

Positive!

Now a change of concavity has occurred, so we can say that,

$(-\frac{2}{3}, -\frac{139}{27})$ is a point of inflexion.

wow thats a great effort, would have taken u forever with latex.

SpiralFlex · Jul 23, 2011

43 minutes 45 seconds.

powlmao · Jul 23, 2011

Spiralflex should be a mod

Hermes1 · Jul 23, 2011

SpiralFlex said:
43 minutes 45 seconds.

is that how long it took u?

fuk thats dedication

you should get like a million rep points for that

SpiralFlex · Jul 23, 2011

arj1 said:
is that how long it took u?

fuk thats dedication

I did 3 hours once... But that's when I wasn't fluent with Latex.

powlmao · Jul 23, 2011

I repped you to much already

You deserve more!!!

SpiralFlex · Jul 23, 2011

Also to OP, I have a few worksheets that may assist you if you want them. Just send a friendly PM. Doing this in Year 10 is certainly impressive.

hscishard · Jul 23, 2011

You spend too much time helping spiralflex. Have you noticed that?

SpiralFlex · Jul 23, 2011

hscishard said:
You spend too much time helping spiralflex. Have you noticed that?

Why?

iRuler · Jul 23, 2011

He's a nice robot, stop discouraging the robot.

hscishard · Jul 23, 2011

Robots can't be discouraged

unless they're programmed to

SpiralFlex · Jul 23, 2011

hscishard said:
Robots can't be discouraged

unless they're programmed to

Quiet human!

Please Help me in my maths problems (1 Viewer)

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