Math Problem Solver

(1) Similarly to typing the keyword sqrt

for expressions involving square roots, expressions involving cubic roots can be typed with use of the keyword root3.

For instance,

a−−√3−b√3

can be entered as

root3(a) - root3(b),

(notice how both a

and b

are enclosed in round brackets).

Another available option is to use raising to the power 1/3

:

a^(1/3) - b^(1/3) .

(2) ∙

As usual, once you have chosen one of the two options described in (1), you should use it throughout when typing an expression involving cubic roots.

∙

Furthermore, if you have decided to use the first option, then any raising to the power of the form m/3,

where m

is an integer number which is not divisible by 3,

should be typed in accordance with the pattern

root3(...)^m

For instance, the expression

a2−−√3

should then be entered as

root3(a)^2

3) As with the previous homework assignment, in accordance with the long-standing traditions,

(a) Irrational numbers of the form

±pq√3,

where p,q∈Z

are integer numbers, must always be (re)written in the form

±ab√3,

where BOTH a

and b

are POSITIVE natural numbers and b

is cubefree (that is, not divisible by any cube of a natural number other than 1

). It is easily seen that a natural number n>1

is cubefree if and only if any prime number in the prime factorization of n

has exponent not exceeding 2.

For example, the irrational number

3−80−−−−√3=−380−−√3=−324⋅5−−−−√3

should be rewritten as

−610−−√3.

(b) Irrational numbers of the form

±pr√3qs√3,

where p,q,r,s∈Z

are integer numbers, should always be (re)written in the form

±ab√3c,

where a,b,c

are POSITIVE natural numbers, b

is cubefree, and the fraction a/c

is in reduced form.

To transform a fraction

±pr√3qs√3

where p,q,r,s∈N+,

to the required form, we can first multiply both the numerator and denominator by s2−−√3,

thereby obtaining the fraction

±pr√3⋅s2−−√3qs√3⋅s2−−√3=±prs2−−−√3qs.

Next, we rewrite rs2−−−√3

in the form xb√3,

where x,b

are positive natural numbers and b

is cubefree, and, finally, transform the fraction

pxqs=ac

where a,c∈N+,

to reduced form, which will give us

±pr√3qs√3=±ab√3c,

as required.

For instance,

4324−−√3=4242−−−√3324−−√3⋅242−−−√3=426⋅32−−−−−√372=169–√372=29–√39.

If one wishes to proceed a bit faster, one can use the canonical prime factorizations in a less straightforward way; e.g.,

418−−√3=42⋅32−−−−√3=4⋅22⋅3−−−−√32⋅32−−−−√3⋅22⋅3−−−−√3=212−−√33,

In cases where a particular prime factorization is hard to come by, one can use suitable online resources (e.g., click on the following link to try one such a resource).

(Limit Rules: Radicals). Consider the limit

limx→49x−20−−−−−−√3−2x+8−−−−−√3x3−64

and let

f(x)=9x−20−−−−−−√3−2x+8−−−−−√3x3−64.

(i) We have that

limx→4(9x−20−−−−−−√3−2x+8−−−−−√3)=

Answer 1 Question 3

0

and

limx→4(x3−64)=

Answer 2 Question 3

0

,

and so the Limit Rules

Multiple choice 1 Question 3

are applicable directly

are not applicable directly

for the evaluation of the limit in question.

(ii) Identify the radical expression (radical) R=R(x)

in the above limit, and enter it in the input field below:

R=

Answer 3 Question 3

root3(9x-20)-root3(2x+8)

.

(iii) 1) Find and enter in the input field below the conjugate radical expression R¯¯¯¯=R¯¯¯¯(x)

:

R¯¯¯¯=

Answer 4 Question 3

root3(9x-20)+root3(2x+8)

.

2) Find also, for the future use, the (exact value of the) limit of the conjugate radical expression as x→4

:

limx→4R¯¯¯¯=

Answer 5 Question 3

.

Prior to entering the value in the input field, please make certain that the value is (re)presented in the form

±ab√3,

where a,b

are positive natural numbers and b

is cubefree (consult the above instructions if necessary).

3) Furthermore, since the limit of the function R¯¯¯¯=R¯¯¯¯(x)

as x→4

is

Multiple choice 2 Question 3

positive

negative

there exists an open interval I

about the point 4

such that the values of the function R¯¯¯¯=R¯¯¯¯(x)

on I

are all

Multiple choice 3 Question 3

positive

negative

(iv) Find the product R⋅R¯¯¯¯

:

R⋅R¯¯¯¯=

Answer 6 Question 3

.

(v) Now assume that x

is an arbitrary point in dom(f)∩I,

where the interval I

has been defined in (iii). Importantly, we then have that x≠4

and that

R¯¯¯¯=R¯¯¯¯(x)≠0

for all x∈dom(f)∩I

(please ensure that you understand that both these facts are true).

Multiply both the numerator and the denominator of the function

f(x)=9x−20−−−−−−√3−2x+8−−−−−√3x3−64

by R¯¯¯¯,

and observe, after applying (iv), that you can cancel the common factor ...

Answer 7 Question 3

,

in the numerator and denominator, thereby obtaining that

f(x)=?1(?2)⋅R¯¯¯¯(3.1)

for all x∈dom(f)∩I,

where

?1=

Answer 8 Question 3

and

?2=

Answer 9 Question 3

.

In other words, if f∗(x)

is the function defined by the formula in the right-hand side of (3.1), we have that

f(x)=f∗(x)

for all x∈dom(f)∩I,

and if the limit of the function f∗(x)

as x→4

exists, so does the limit of the function f(x)

as x→4,

and both the limits are equal.

(vi) By your results in (v), the Limit Rules

Multiple choice 4 Question 3

are applicable

are not applicable

for the evaluation of the limit of the function f∗(x)

as x→4,

and hence

L=limx→4f(x)=limx→4f∗(x)=

Answer 10 Question 3

.

(enter an asterisk in the input field, if the limit does not exist).

Prior to entering the value in the input field, please make certain that the value is (re)presented in the form

±ab√3/c,

where a,b,c

are positive natural numbers, b

is cubefree, and the fraction a/c

is in reduced form (consult the above instructions if necessary).

(vii) Finally, if the limit L

in question exists, round it to 5 decimal places (enter an asterisk otherwise):

L≈

Answer 11 Question 3

.