Finding Infliction points and the equations of their slope

[elfy]

Homework Statement

x^4 - 6x^3 + 12x^2 - 8x

a) Find all points of infliction of this function
b) Sketch the function and identify all convex and concave portions of the curve
c) Find the equations of the slopes at each point of infliction

Homework Equations

x^4 - 6x^3 + 12x^2 - 8x

The Attempt at a Solution

Hello,
What I attempted to do was to do the second derivative of the function (as I don't need to determine min/max there is no need to bother with the first derivative right, i.e. finding the points at which the slope is 0?). After finding the second derivative I just solve it such as that y''=0 for a given x.

a)
y' = 4x^3 - 18x^2 + 24x - 8
y'' = 12x^2 - 36x + 24

Solving X such as that y'' = 0 (infliction points):
[-(-36) +/- sqrt(-36^2 - (4*12*24)] / 2*12

Which yields: (36+/- 12) / 24 ----> x = 2, x = 1

So the infliction points are x=2 and x=1.

b)
I think I got this one right as it's just a matter of plugging in different values for X and plot it in a graph.

c)
This is the one I can't figure out how to do. I am supposed to find the equations of the slopes at each point of infliction, but I don't quite understand the question. Do I just plug in the x value in the original function?
Ex. x = 1 ---> Equation at infliction pt: 1-6+12-8 = -1 ??

I would really appreciate some guidance as to how I am supposed to do c).

Thanks!

 

[mstud]

Homework Statement

x^4 - 6x^3 + 12x^2 - 8x

a) Find all points of infliction of this function
b) Sketch the function and identify all convex and concave portions of the curve
c) Find the equations of the slopes at each point of infliction

Homework Equations

x^4 - 6x^3 + 12x^2 - 8x

The Attempt at a Solution

Hello,
What I attempted to do was to do the second derivative of the function (as I don't need to determine min/max there is no need to bother with the first derivative right, i.e. finding the points at which the slope is 0?). After finding the second derivative I just solve it such as that y''=0 for a given x.

a)
y' = 4x^3 - 18x^2 + 24x - 8
y'' = 12x^2 - 36x + 24

Solving X such as that y'' = 0 (infliction points):
[-(-36) +/- sqrt(-36^2 - (4*12*24)] / 2*12

Which yields: (36+/- 12) / 24 ----> x = 2, x = 1

So the infliction points are x=2 and x=1.

b)
I think I got this one right as it's just a matter of plugging in different values for X and plot it in a graph.

c)
This is the one I can't figure out how to do. I am supposed to find the equations of the slopes at each point of infliction, but I don't quite understand the question. Do I just plug in the x value in the original function?
Ex. x = 1 ---> Equation at infliction pt: 1-6+12-8 = -1 ??

I would really appreciate some guidance as to how I am supposed to do c).

Thanks!

Take a look here: http://en.wikipedia.org/wiki/Slope#Calculus"

The slope of a non-linear function is given by the derivative of the function in a certain point, so plug the x-values into the y' you found

Hope you're helped by this !

You were asked to find the equation of the slope, therefor you might as well take a look here:http://en.wikipedia.org/wiki/Tangent#More_rigorous_description"

The last link describes how to find the equation when you know the x and y values of the point, and the slope of the function's tangent in the point.

Then you will 've reached your goal... I guess

 

[elfy]

Take a look here: http://http://en.wikipedia.org/wiki/Slope#Calculus"

The slope of a non-linear function is given by the derivative of the function in a certain point, so plug the x-values into the y' you found

Hope you're helped by this !

Thanks a bunch mate! That definitely helped!
But how is that "an equation" as y' will equal 0 at x=2 and equal 2 at x=1? Isn't that just numbers ;)

I thought you had to find some type of equation in the sense (x + y...). English is not my first language so sometimes when I'm doing maths in english I get confused :)

Again, thank you very much!

 

[mstud]

You should find an equation, did you see I edited my first post, because what I first thought was that you should find the slope, and not the equation of the slope... Sorry , I should rather have made a new post

To find the equation when having x-value in the point, y-value in the point (found by plugging into y), and the "value" of the slope y'(x), you use:

y="y-value"+ "slope value" times (x-"x-value")

Again, sorry for the inconvenience...

It took me a second more to understand your question fully than to press the "submit reply" button... since English is not my first language either :)

 

[elfy]

hey mstud - no worries at all! Now I learned 2 new things even though I just asked for 1 ;)

Ok, after reading your post and the wikipedia page i tried to solve it:

x = 2 (infliction point found earlier)

y = f(2) + f'(2)(x-2)

--f(2) = 2^4 -6(2^3) + 12(2^2) - (8*2) = 0
--f'(2) = 4(2^3) - 18(2^2) + (24*2) - 8 = 4

Plugging that in the formula yields:

y = 0 + 4(x-2) ----> y = 4x - 8

Solving for x yields x = (y-8)/4

Is this the right way of doing it or have I got it wrong once again :)

 

[mstud]

hey mstud - no worries at all! Now I learned 2 new things even though I just asked for 1 ;)

Ok, after reading your post and the wikipedia page i tried to solve it:

x = 2 (infliction point found earlier)

y = f(2) + f'(2)(x-2)

--f(2) = 2^4 -6(2^3) + 12(2^2) - (8*2) = 0
--f'(2) = 4(2^3) - 18(2^2) + (24*2) - 8 = 4

Plugging that in the formula yields:

y = 0 + 4(x-2) ----> y = 4x - 8

Solving for x yields x = (y-8)/4

Is this the right way of doing it or have I got it wrong once again :)

The answer is y = 4x - 8 , just like you found. It is not necessary to solve for x, but if you should have done so, it would have been:

$$y=4x-8 \Leftrightarrow y+8=4x \Leftrightarrow x= (y+8)/4$$

Nice work as far as you had to go, and I just wanted to comment your solving for x so you would know till next time :)

 

[elfy]

oh yeah, that was a silly mistake by me, should have seen it's +8 and not - hehe.

Thank you very much for your time and effort to help me out mate, I really appreciate it!

 

[mstud]

Such mistakes we all make, sometimes.

You're welcome!