Questions related to Relations and Functions

[Raghav Gupta]

Homework Statement

1. Range of the function $\sqrt {x^2+x+1}$ is equal to?

2.ƒ:R---->R is defined as ƒ(x) = x² -3x +4, then f ^-1 (2) is equal to?

Homework Equations

NA

The Attempt at a Solution

For the first one tried squaring on both the sides but that does not give linear x in terms of y for finding the range.

For second one, I can directly substitute f(x) as 2 for getting the answer.
But I have a confusion that a function must be injective and surjective for the inverse otherwise it inverse must not exist.

 

[fourier jr]

1. you must have $x^2 + x + 1 \geq 0$ so for what x is that true?

2. I think you could start with $2 = x^2 - 3x + 4$ & go from there

 

[Raghav Gupta]

1. you must have $x^2 + x + 1 \geq 0$ so for what x is that true?

2. I think you could start with $2 = x^2 - 3x + 4$ & go from there

1. I'm not asking for domain but range.
I know that the domain would be set of all real numbers.

2. That I know but I think f inverse should not exist as the function is not one one and onto?

 

[LCKurtz]

1. you must have $x^2 + x + 1 \geq 0$ so for what x is that true?

2. I think you could start with $2 = x^2 - 3x + 4$ & go from there

1. I'm not asking for domain but range.
I know that the domain would be set of all real numbers.

You have a square root to worry about, so not every $x$ works in the domain. And what $x$'s work determine the range.

2. That I know but I think f inverse should not exist as the function is not one one and onto?

Sometimes when $f^{-1}$ does not exist, the notation such as $f^{-1}(2)$ means the set of all $x$ such that $f(x)=2$.

 

[Raghav Gupta]

You have a square root to worry about, so not every $x$ works in the domain. And what $x$'s work determine the range.

But I see here that if we choose any x here in this case , the value of
x² + x + 1 is always greater than zero,
so not to worry for square root in this case.

 

[LCKurtz]

But I see here that if we choose any x here in this case , the value of
x² + x + 1 is always greater than zero,
so not to worry for square root in this case.

That's right, but that does not mean the range is $(0,\infty)$. For example, what $x$ gives $f(x) = 1/2$?

 

[Raghav Gupta]

That's right, but that does not mean the range is $(0,\infty)$. For example, what $x$ gives $f(x) = 1/2$?

No x gives that value

 

[LCKurtz]

No x gives that value

Right. So you still have to figure out the range.

 

[Raghav Gupta]

Right. So you still have to figure out the range.

Ya, the range then must be greater then 1/2 also.
What is the method to find the particular value?

 

[LCKurtz]

Ya, the range then must be greater then 1/2 also.
What is the method to find the particular value?

You have to figure out the least that $x^2+x+1$, and hence its square root, can be.

 

[Raghav Gupta]

You have to figure out the least that $x^2+x+1$, and hence its square root, can be.

I differentiated it,
Got 2x + 1 = 0
Hence x = -1/2, a minima.
Substituting in function we get square root of 3/4 which is √3/2
Hence range is [√3/2,∞)
Thanks.
The word least provoked the differentiation.

 

[LCKurtz]

Good. Note that you could have also found the min value without calculus by completing the square.

 

[Raghav Gupta]

Good. Note that you could have also found the min value without calculus by completing the square.

Hmm that's also fine.
As the thread is related to relations and functions.
I wanted to ask only a last question.

If $f(x) = sin^2x + sin^2(x+ π/3) + cosxcos(x+ π/3)$ and $g(5/4)=1$, then $(gof)(x)$ is equal to?
Options are
0
1
sinx
None of these

I know gof(x) is g(f(x)) but here g(x) is not given.

 

[Raghav Gupta]

Anybody there or should I start a new thread?

 

[SammyS]

Anybody there or should I start a new thread?

I say yes - new thread. It's quite a different question.

Composition in LaTeX is $(f\circ g)(x) \ \ \ \leftarrow\ \ \ \text{(f\circ g)(x)}$ .

 

[Mark44]

Anybody there or should I start a new thread?

Yes, new thread, which you already did.

See my reply in that thread.