Find all three solutions for the given equation.
Answers
Answer:
Solving epsilon-delta problems
Math 1A, 313,315 DIS
September 29, 2014
There will probably be at least one epsilon-delta problem on the midterm and the final.
These kind of problems ask you to show1
that
limx→a
f(x) = L
for some particular f and particular L, using the actual definition of limits in terms of ’s
and δ’s rather than the limit laws. For example, there might be a question asking you to
show that
limx→a
7x + 3 = 7a + 3 (1)
or
limx→5
x
2 − x − 1 = 19, (2)
using the definition of a limit.
1 The rules of the game
Normally, the answer to this kind of question will be of the following form:
Given > 0, let δ = [something positive, usually depending on and a]. If
0 < |x − a| < δ then [some series of steps goes here], so |f(x) − L| < .
Some examples of this are Examples 2-4 of section 2.4. Note that “[some series of steps goes
here]” should consist of a proof that |f(x) − L| < , from the assumptions that
• > 0
• δ is whatever we said it was, and
• 0 < |x − a| < δ.
1
i.e., prove
1 In these kind of problems, much of the work goes into figuring out what δ should be.
None of this work is shown in the actual answer. To clarify: in examples 2-4 of section 2.4,
each “solution” consists of two parts. Part 2 (“showing that this δ works”) is the actual
answer–what you would turn in if asked this question on a homework or an exam. Part 1
(“guessing a value for δ”) is the bulk of the work done to produce this answer.
So there’s a sense in which you don’t have to show your work in this kind of problem; it
suffices to just write down the final answer. This is a little strange because for most math
problems it is necessary to show your work. For example, if there was a problem asking you
to evaluate
limx→1
x
4 − 1
x − 1
,
it would not be acceptable to just write down “4.” This would be unacceptable because
there’s no way for the person reading your answer to see why the limit should be 4. But if
the answer to a question is a proof, rather than a number or an expression, then the reader
can see directly whether or not the answer is correct, because the correctness of a proof is
self-evident. In problems where the answer is a number or an expression, when we say “show
your work” we really mean “show that the answer is correct.” For example, a more correct
answer to limx→1(x
4 − 1)/(x − 1) would be
limx→1
x
4 − 1
x − 1
= limx→1
(x
3 + x
2 + x + 1)(x − 1)
x − 1
= limx→1
(x
3 + x
2 + x + 1)
= 13 + 12 + 1 + 1 = 4.
The first step is just rewriting the thing whose limit is being taken. The second step is using
the fact that limx→1 only looks at values of x that aren’t 1, for which we can cancel out the
factors of (x − 1). The third step is the direct substitution principle for polynomials, and
the last step is basic arithmetic.
2 Common mistakes
From looking through people’s homework, I got the impression that the following mistakes
were common:
• Dividing by zero, or treating ∞ as if it were an actual number.
• Writing things like
limx→1
x
4 − 1
x − 1
= x
3 + x
2 + x + 1 = 4.
In limx→1
x
4−1
x−1
, the variable x is a bound variable. To paraphrase Wikipedia, “there
is nothing called x on which limx→1(x
4 − 1)/(x − 1) could depend.” It doesn’t make
sense to say that the limit is equal to x
3 + x
2 + x + 1, because what is x?
• Not specifying what you chose δ to be! If you don’t do this, it’s really unclear what
you’re ultimately trying to prove.
2• Confusing the preliminary analysis to figure out δ, with the actual answer (the proof),
or flat out omitting the actual answer.
Answer :
x = -1 , (1 + √3i)/2 , (1 - √3i)/2
Solution :
- Given : x³ = -1
- To find : Solutions (Roots) of the given equation x³ = -1
We have ,
=> x³ = -1
=> x³ + 1 = 0
=> x³ + 1³ = 0
=> (x + 1)(x² - x•1 + 1²) = 0
=> (x + 1)(x² - x + 1) = 0
Here ,
Two cases arises ;
• x + 1 = 0 OR
• x² - x + 1 = 0
•Case1 :
If x + 1 = 0 , then x = -1
• Case2 :
If x² - x + 1 = 0 , then
This is a quadratic equation in x .
Now ,
Comparing the equation x² - x + 1 = 0 with the general quadratic equation ax² + bx + c = 0 , we have ;
a = 1
b = -1
c = 1
Now ,
The discriminant will be given as ;
=> D = b² - 4ac
=> D = (-1)² - 4•1•1
=> D = 1 - 4
=> D = -3
=> D < 0 , thus the equation x² - x + 1 = 0 will have complex conjugate pair of roots .
Now ,
The roots will be given as ;
x = ( -b ± √D)/2a
x = [-(-1) ± √(-3)]/2•1
x = (1 ± √3i)/2
x = (1 + √3i)/2 , (1 - √3i)/2
Hence ,
All the three roots are ;
x = -1 , (1 + √3i)/2 , (1 - √3i)/2