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#1 2021-05-13 14:57:53

mathland
Member
Registered: 2021-03-25
Posts: 444

Domain of Cube Root Function

Find the domain of f(t) = cuberoot{ t + 4}.

Let me see.

I must set the radicand to be greater than or equal to 0 and solve for t. Yes?

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#2 2021-05-13 19:16:22

Bob
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Registered: 2010-06-20
Posts: 10,058

Re: Domain of Cube Root Function

if the question involved a square root I would agree.

But cuberoot(a negative) is possible without diving into complex numbers

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2021-05-13 19:22:03

mathland
Member
Registered: 2021-03-25
Posts: 444

Re: Domain of Cube Root Function

Bob wrote:

if the question involved a square root I would agree.

But cuberoot(a negative) is possible without diving into complex numbers

Bob

Do I set the radicand to be greater than or equal to 0?

t + 4 >= 0

t >= -4...Domain?

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#4 2021-05-13 19:45:36

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Domain of Cube Root Function

I've just tried the graph and this confirms your answer.

But .... am I making a silly mistake here?:

Let t = -12

f(-12) = cuberoot (-12 +4) = cuberoot(-8) = -2

Instead of using the function grapher I tried the equation grapher at

https://www.mathsisfun.com/data/grapher-equation.html  and used the equation y^3 = x + 4

This gives the complete graph including values of y below zero.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#5 2021-05-13 20:40:13

mathland
Member
Registered: 2021-03-25
Posts: 444

Re: Domain of Cube Root Function

Bob wrote:

I've just tried the graph and this confirms your answer.

But .... am I making a silly mistake here?:

Let t = -12

f(-12) = cuberoot (-12 +4) = cuberoot(-8) = -2

Instead of using the function grapher I tried the equation grapher at

https://www.mathsisfun.com/data/grapher-equation.html  and used the equation y^3 = x + 4

This gives the complete graph including values of y below zero.

Bob

The range is [-3, 3].

Ever tried demos? It's a free online graphing calculator.

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#6 2021-05-14 01:12:34

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Domain of Cube Root Function

I'm still thinking about the range.  Maybe more later.

In fact now:

Let t = 121 then f = cuberoot(121 + 4) = cuberoot 125 = 5 > 3
Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#7 2021-05-14 06:19:43

mathland
Member
Registered: 2021-03-25
Posts: 444

Re: Domain of Cube Root Function

Bob wrote:

I'm still thinking about the range.  Maybe more later.

In fact now:

Let t = 121 then f = cuberoot(121 + 4) = cuberoot 125 = 5 > 3
Bob

I'll wait to discuss the range further when you have time.

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#8 2021-05-14 08:32:49

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: Domain of Cube Root Function

I think both the domain and the range are (-∞,∞)

First the domain.

All negatives real numbers have a negative, real cube root.

Say n is the cube root of t, n and t > 0, then -n is the cube root of -t. So all negative values of (t+4) have a cube root.

Range.  f = cuberoot(t+4) is symmetrical about the point (-4,0)

proof:  Suppose f = cuberoot(t+4) for t > -4

Then the vector from [-4,0] to [t , f] is [t+4, f]

Now consider the point [-t - 8, cuberoot(-t-8+4)]

The vector from this point to [-4,0] is [-4 -(-t-8) , -cuberoot(-t-4)] = [t + 4, +cuberoot(t+4)] = [t+4,f]

These two vectors are the same showing the required symmetry.

There is no upper bound to f.

Proof.  Suppose N is the upper bound.

Form the number t = (N+1)^3 - 4

f(t) = cuberoot[(N+1)^3 - 4 + 4] = cuberoot[(N+1)^3] = N + 1 which is above the supposed upper bound.  So there is no upper 
bound.  So the range extends to + ∞.  But the graph has rotational symmetry. So the lower limit of the range is -∞.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#9 2021-05-14 11:07:54

mathland
Member
Registered: 2021-03-25
Posts: 444

Re: Domain of Cube Root Function

Bob wrote:

I think both the domain and the range are (-∞,∞)

First the domain.

All negatives real numbers have a negative, real cube root.

Say n is the cube root of t, n and t > 0, then -n is the cube root of -t. So all negative values of (t+4) have a cube root.

Range.  f = cuberoot(t+4) is symmetrical about the point (-4,0)

proof:  Suppose f = cuberoot(t+4) for t > -4

Then the vector from [-4,0] to [t , f] is [t+4, f]

Now consider the point [-t - 8, cuberoot(-t-8+4)]

The vector from this point to [-4,0] is [-4 -(-t-8) , -cuberoot(-t-4)] = [t + 4, +cuberoot(t+4)] = [t+4,f]

These two vectors are the same showing the required symmetry.

There is no upper bound to f.

Proof.  Suppose N is the upper bound.

Form the number t = (N+1)^3 - 4

f(t) = cuberoot[(N+1)^3 - 4 + 4] = cuberoot[(N+1)^3] = N + 1 which is above the supposed upper bound.  So there is no upper 
bound.  So the range extends to + ∞.  But the graph has rotational symmetry. So the lower limit of the range is -∞.

Bob

Great math notes as always. I will explore this question more and more when time allows.

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#10 2021-05-14 22:03:23

Bob
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Registered: 2010-06-20
Posts: 10,058

Re: Domain of Cube Root Function

Further notes on this:

cuberoot(t+4) is cuberoot(t) translated 4 units left in a horizontal direction.

Examine y = (x)^(1/3)

This graph and its inverse are 1:1 functions  so we can proceed  to raise both sides to the power 3

y^3 = x

We can reflect the graph of this function in the line y=x by swapping the x and y terms.

y = x^3

We know all about this cubic already.  Both its domain and range are (-∞,∞)

The reflection interchanges  domain and range.  In other examples this might matter but as both are  (-∞,∞) it doesn't matter.

The translation might effect some domains, but again, not this one.

So we can get domain and range just by considering simple transformations of the graph.

General hint.  Cubics in general have certain common properties. We can call all such curves a 'family of curves' and use general properties to learn about new members.

Cubics with a positive x^3 term all start in the 3rd quadrant and finish in the 1st. If x^3 is negative this becomes 2nd and 4th quadrants.

Cubics have two stationary points (max and min) unless the two points merge into one in which case we have a point of inflexion.

Here are some general properties for quadratics:

All have a line of symmetry  x = -b/2a .  This line goes through the stationary point which is a minimum when the x^2 term is positive and a maximum when the x^2 term is negative.

Graphs of the form y = a^x

These all go through (0,1)

The gradient function is k . a^x

When k = 1, we use 'e' , making a function that has the interesting property that dy/dx = y.

I'm sure I did a thread on this but I cannot find it now using search.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#11 2021-05-15 02:22:00

mathland
Member
Registered: 2021-03-25
Posts: 444

Re: Domain of Cube Root Function

Bob wrote:

Further notes on this:

cuberoot(t+4) is cuberoot(t) translated 4 units left in a horizontal direction.

Examine y = (x)^(1/3)

This graph and its inverse are 1:1 functions  so we can proceed  to raise both sides to the power 3

y^3 = x

We can reflect the graph of this function in the line y=x by swapping the x and y terms.

y = x^3

We know all about this cubic already.  Both its domain and range are (-∞,∞)

The reflection interchanges  domain and range.  In other examples this might matter but as both are  (-∞,∞) it doesn't matter.

The translation might effect some domains, but again, not this one.

So we can get domain and range just by considering simple transformations of the graph.

General hint.  Cubics in general have certain common properties. We can call all such curves a 'family of curves' and use general properties to learn about new members.

Cubics with a positive x^3 term all start in the 3rd quadrant and finish in the 1st. If x^3 is negative this becomes 2nd and 4th quadrants.

Cubics have two stationary points (max and min) unless the two points merge into one in which case we have a point of inflexion.

Here are some general properties for quadratics:

All have a line of symmetry  x = -b/2a .  This line goes through the stationary point which is a minimum when the x^2 term is positive and a maximum when the x^2 term is negative.

Graphs of the form y = a^x

These all go through (0,1)

The gradient function is k . a^x

When k = 1, we use 'e' , making a function that has the interesting property that dy/dx = y.

I'm sure I did a thread on this but I cannot find it now using search.

Bob

Great notes. I will look further into this information and get back to you, if need be. More math questions posted later on today and tomorrow, Lord willing.

Last edited by mathland (2021-05-15 02:22:25)

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