Find No Trivial Solution for Ax 0 Calculator

What's this about?

This is, in some way, the opposite of curve sketching. Curve sketching means you got a function and are looking for roots, turning and inflection points. What we do here is the opposite: Your got some roots, inflection points, turning points etc. and are looking for a function having those.

How to reconstruct a function?

Primarily, you have to find equations and solve them. This gives you the coefficients of your function. Here is an example: Let's assume we are looking for a function of degree having a minimum turning point at (1|-4) and a maximum turning point at (-1|3).

You are looking for a function with:
quadratic function
Maximum turning point at (-1|3)
Minimum turning point at (1|-4)

Mathepower found the following function:
1*f(x)=1*x^2+-3.5*x+-1.5
This is the graph of your function.

Roots at -0.386; 3.886
y-axis intercept at (0|-1.5)
Maximum and minimum turning points at (1.75|-4.563)
Inflection points

This is how Mathepower calculated:
f(x)=1*a*x^2+1*b*x+1*c

The point at (-1|3) gives the equation :
f(-1)=3
1*a*-1^2+1*b*-1+1*c=3
simplified: :
1a-1b+1c=3

The point at (1|-4) gives the equation :
f(1)=-4
1*a*1^2+1*b*1+1*c=-4
simplified: :
1a+1b+1c=-4

So, we got the following system of equations: :

a	-1b	+c	=	3
a	+b	+c	=	-4

This is how to solve this system of equations:

a	-1b	+c	=	3
a	+b	+c	=	-4

a	-1b	+c	=	3
	2b		=	-7

( -1 times line 1 was added to line 2 )

a	-1b	+c	=	3
	b		=	-3,5

( The 2 line was divided by 2 )

2 line:	b+0c = -3,5
c can be chosen freely
Solve for b : :	b = 0c -3,5

1 line:

-1b

Substitute variables already known:

-1⋅(0c -3,5)

+⋅1c

Solve for a :

a = -1c -0,5

Set a equal to
This means that c is equal to -1,5

Inserting shows that the function equals to ist.

How to find a function through given points?

The general rule is that for any n given points there is a function of degree n-1 whose graph goes through them. So e.g. you find by solving equations a function of degree through the four points (-1|3), (0|2), (1|1) und (2|4):

You are looking for a function with:
Function of degree 3
Point at (-1|3)
Point at (0|2)
Point at (1|1)
Point at (2|4)

Mathepower found the following function:
1*f(x)=0.667*x^3+-1.667*x+2
This is the graph of your function.

Roots at -2
y-axis intercept at (0|2)
Maximum and minimum turning points at (-0.913|3.014); (0.913|0.986)
Inflection points at (0|2)

This is how Mathepower calculated:
f(x)=1*a*x^3+1*b*x^2+1*c*x+1*d

The point at (-1|3) gives the equation :
f(-1)=3
1*a*-1^3+1*b*-1^2+1*c*-1+1*d=3
simplified: :
-1a+1b-1c+1d=3

The point at (0|2) gives the equation :
f(0)=2
1*a*0^3+1*b*0^2+1*c*0+1*d=2
simplified: :
0a+0b+0c+1d=2

The point at (1|1) gives the equation :
f(1)=1
1*a*1^3+1*b*1^2+1*c*1+1*d=1
simplified: :
1a+1b+1c+1d=1

The point at (2|4) gives the equation :
f(2)=4
1*a*2^3+1*b*2^2+1*c*2+1*d=4
simplified: :
8a+4b+2c+1d=4

So, we got the following system of equations: :

-1a	+b	-1c	+d	=	3
			d	=	2
a	+b	+c	+d	=	1
8a	+4b	+2c	+d	=	4

This is how to solve this system of equations:

-1a	+b	-1c	+d	=	3
			d	=	2
a	+b	+c	+d	=	1
8a	+4b	+2c	+d	=	4

-1a	+b	-1c	+d	=	3
			d	=	2
a	+b	+c	+d	=	1
	-4b	-6c	-7d	=	-4

( -8 times line 3 was added to line 4 )

-1a	+b	-1c	+d	=	3
			d	=	2
	2b		+2d	=	4
	-4b	-6c	-7d	=	-4

( 1 times line 1 was added to line 3 )

a	-1b	+c	-1d	=	-3
			d	=	2
	2b		+2d	=	4
	-4b	-6c	-7d	=	-4

( The 1 line was divided by -1 )

a	-1b	+c	-1d	=	-3
			d	=	2
	2b		+2d	=	4
		-6c	-3d	=	4

( 2 times line 3 was added to line 4 )

a	-1b	+c	-1d	=	-3
	2b		+2d	=	4
			d	=	2
		-6c	-3d	=	4

( the 3 line was interchanged with the 2 line )

a	-1b	+c	-1d	=	-3
	b		+d	=	2
			d	=	2
		-6c	-3d	=	4

( The 2 line was divided by 2 )

a	-1b	+c	-1d	=	-3
	b		+d	=	2
		-6c	-3d	=	4
			d	=	2

( the 4 line was interchanged with the 3 line )

a	-1b	+c	-1d	=	-3
	b		+d	=	2
		c	+0,5d	=	-0,667
			d	=	2

( The 3 line was divided by -6 )

3 line:

+0,5d

-0,667

Substitute variables already known:

+0,5⋅2

-0,667

Solve for c :

c = -1,667

2 line:

Substitute variables already known:

+⋅2

Solve for b :

b = 0

1 line:

-1b

-1d

-3

Substitute variables already known:

-1⋅0

+⋅(-1,667)

-1⋅2

-3

Solve for a :

a = 0,667

Inserting shows that the function equals to ist.

How to find a function with a given inflection point?

An inflection point gives multiple equations: On the one hand, you got the y-value. On the other hand, you know that the second derivative is at an inflection point. Let's take a look at an example for a function of degree having an inflection point at (1|3):

You are looking for a function with:
Function of degree 3
root at 2
root at 4
Inflection point at (1|3)

Mathepower found the following function:
1*f(x)=1*c*x+1*d
This is the graph of your function.

Roots
y-axis intercept at (0|0)
Maximum and minimum turning points
Inflection points

This is how Mathepower calculated:
f(x)=1*a*x^3+1*b*x^2+1*c*x+1*d

The point at (1|3) gives the equation :
f(1)=3
1*a*1^3+1*b*1^2+1*c*1+1*d=3
simplified: :
1a+1b+1c+1d=3

So, we got the following system of equations: :

This is how to solve this system of equations:

1 line:	c+1d = 3
d can be chosen freely
Solve for c : :	c = -1d +3

Inserting shows that the function equals to ist.

And how to use that in my example?

Just enter your exercise above. Mathepower shows how it works by doing a free step-by-step calculation. Or just make up any interesting exercise and check out what Mathepower does.

Find No Trivial Solution for Ax 0 Calculator

Source: https://www.mathepower.com/en/findingfunctions.php

Clark Forges

Find No Trivial Solution for Ax 0 Calculator

What's this about?

How to reconstruct a function?

How to find a function through given points?

How to find a function with a given inflection point?

And how to use that in my example?

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