MohammedPatel
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#1
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Hi.

I am having some disagreements with the answers provided by my textbook and would really like some help with where I might have gone wrong.

Questions from textbook:

1) Determine which of these mappings are functions:
e) x -> x

I believe this to not be a function as a square root of a number has a positive and negative root. Eg. if x = 4 then
x can equal +2 or -2. However the textbook says its a 'function when x
≥ 0'

4) If f(x) = the value of x correct to the nearest integer, find f (-3.5)

The textbook answer is 'not defined' - which I assume is because the answer could be -3 or -4. But I don't understand, if there are two possible outcomes why is classed as a function as appose to just a mapping?

5) If f(x) = sin x, find f(½ π) and f(2/3 π)

I went about this by replacing x with 1/2 π and 2/3 π
So for example in my calculator I typed: sin (1/2 x π)
And the answers I got were
f(½ π) = 0.0274 and f(2/3 π) = 0.0365
where as the textbook says the answers are 1, and
3 /2 respectively.

I believe I've gone horribly wrong with question 5, and would really appreciate the help. As I am studying independently and sitting the exams as a private candidate I do not have a teacher I can ask for help.
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davros
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(Original post by MohammedPatel)
Hi.

I am having some disagreements with the answers provided by my textbook and would really like some help with where I might have gone wrong.

Questions from textbook:



I believe this to not be a function as a square root of a number has a positive and negative root. Eg. if x = 4 then
x can equal +2 or -2. However the textbook says its a 'function when x
≥ 0'


The textbook answer is 'not defined' - which I assume is because the answer could be -3 or -4. But I don't understand, if there are two possible outcomes why is classed as a function as appose to just a mapping?

[FONT=arial]

I went about this by replacing x with 1/2 π and 2/3 π
So for example in my calculator I typed: sin (1/2 x π)
and the answers i got were
f(½ π) = 0.0274 and f(2/3 π) = 0.0365
where as the textbook says the answers are 1, and
3 /2 respectively.

I believe I've gone horribly wrong with question 5, and would really appreciate the help. As I am studying independently and sitting the exams as a private candidate I do not have a teacher I can ask for help.
For Q1 the square root symbol denotes the positive square root, so your book is correct - the function is only defined when x >= 0.

For Q5 you just need to put your calculuator in radians mode then you'll get the correct answer
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joostan
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(Original post by MohammedPatel)
Hi.

I am having some disagreements with the answers provided by my textbook and would really like some help with where I might have gone wrong.

Questions from textbook:



I believe this to not be a function as a square root of a number has a positive and negative root. Eg. if x = 4 then
x can equal +2 or -2. However the textbook says its a 'function when x
≥ 0'


The textbook answer is 'not defined' - which I assume is because the answer could be -3 or -4. But I don't understand, if there are two possible outcomes why is classed as a function as appose to just a mapping?

[FONT=arial]

I went about this by replacing x with 1/2 π and 2/3 π
So for example in my calculator I typed: sin (1/2 x π)
and the answers i got were
f(½ π) = 0.0274 and f(2/3 π) = 0.0365
where as the textbook says the answers are 1, and
3 /2 respectively.

I believe I've gone horribly wrong with question 5, and would really appreciate the help. As I am studying independently and sitting the exams as a private candidate I do not have a teacher I can ask for help.
For e) the function \sqrt{x} is generally taken to be exclusively positive, unless you stick a \pm in front, hence is a function.

For the last you need to be in radians, not in degrees on your calculator.

As for Q4 it seems that it was never defined as a function in the first place, as it is not well defined.
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MohammedPatel
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(Original post by davros)
For Q1 the square root symbol denotes the positive square root, so your book is correct - the function is only defined when x >= 0.

For Q5 you just need to put your calculuator in radians mode then you'll get the correct answer
Thanks!
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MohammedPatel
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(Original post by joostan)

As for Q4 it seems that it was never defined as a function in the first place, as it is not well defined.
Thanks for the help with the other two questions, but Q4 is defined as a function in my textbook. Its the answer to f(-3.5) that's 'not defined'. Also isn't anything that starts with f(x) confirmed to be a function?
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davros
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(Original post by MohammedPatel)
Thanks for the help with the other two questions, but Q4 is defined as a function in my textbook. Its the answer to f(-3.5) that's 'not defined'. Also isn't anything that starts with f(x) confirmed to be a function?
That's just a piece of notation. It is suggestive of a function, yes, but don't read too much into it!

In this case f(x) can be a function on a restricted domain of values - f(4) makes sense as does f(3.8). Can you see why f(-3.5) isn't defined and why f(-2.5) wouldn't be defined either?
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MohammedPatel
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(Original post by davros)
That's just a piece of notation. It is suggestive of a function, yes, but don't read too much into it!

In this case f(x) can be a function on a restricted domain of values - f(4) makes sense as does f(3.8). Can you see why f(-3.5) isn't defined and why f(-2.5) wouldn't be defined either?
Ah, that makes sense. Thanks.
You know how for question 3 the answer is 'Function, when x 0', in a similar manner could it be said that: 'f(x) = the value of x to the nearest integer, is a function when x doesn't end in .5'?
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davros
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(Original post by MohammedPatel)
Ah, that makes sense. Thanks.
You know how for question 3 the answer is 'Function, when x 0', in a similar manner could it be said that: 'f(x) = the value of x to the nearest integer, is a function when x doesn't end in .5'?
Exactly! It's a bit of an awkward domain to describe, although there are ways you could try to put it in symbols.
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