Another function

**thereddevils** · March 4th 2009, 03:48 AM

Find the range of each function . For any function that is not one to one , give two distinct values of x which have the same image .

(1) $f(x)=sin x$ , x are real numbers

(2) $g(x)=\sqrt{x}+\sqrt{1+x}$ , $o\geq x \geq1$ ,

(3) $h(x)=In{x}$ , $x>0$

the range for sine function is [-1,1]

but for the rest , i am not really sure .

**Grandad** · March 4th 2009, 05:13 AM

Hello thereddevils

Originally Posted by thereddevils

Find the range of each function . For any function that is not one to one , give two distinct values of x which have the same image .

(1) $f(x)=sin x$ , x are real numbers

(2) $g(x)=\sqrt{x}+\sqrt{1+x}$ , $o\geq x \geq1$ ,

(3) $h(x)=In{x}$ , $x>0$

the range for sine function is [-1,1]

but for the rest , i am not really sure .

(1) $f$ is not one-to-one; e.g. $f(0) = f(\pi)$

(2) I presume you mean $0\le x\le 1$ , in which case, since $g$ is an increasing function for $x \ge 0$ , the range is $[1, 1+\sqrt{2}]$ ; and $g$ is one-to-one.

(3) The function $\text{ln}x$ , for $x > 0$ , has the whole of $\mathbb{R}$ as its range. And it is also one-to-one.

Grandad

**HallsofIvy** · March 4th 2009, 05:20 AM

Originally Posted by thereddevils

Find the range of each function . For any function that is not one to one , give two distinct values of x which have the same image .

(1) $f(x)=sin x$ , x are real numbers

(2) $g(x)=\sqrt{x}+\sqrt{1+x}$ , $o\geq x \geq1$ ,

(3) $h(x)=In{x}$ , $x>0$

the range for sine function is [-1,1]

but for the rest , i am not really sure .

(1) You know that sin x is periodic with period $2\pi$ . That should give you an easy answer.
(2) This is a little harder because this is one to one. I would show that by arguing that the derivative is always positive and so the function is increasing but I don't know what methods you have available.
(3) I don't recognise "In(x)". Do you mean the natural logarighm? That is ln(x)- "logarithm" starts with a "l" not "I"! If so, same comments as in (2) apply.

**thereddevils** · March 4th 2009, 05:39 AM

Originally Posted by Grandad

Hello thereddevils(1) $f$ is not one-to-one; e.g. $f(0) = f(\pi)$

(2) I presume you mean $0\le x\le 1$ , in which case, since $g$ is an increasing function for $x \ge 0$ , the range is $[1, 1+\sqrt{2}]$ ; and $g$ is one-to-one.

(3) The function $\text{ln}x$ , for $x > 0$ , has the whole of $\mathbb{R}$ as its range. And it is also one-to-one.

Grandad

Thanks once again . For (2) If the function is one to one , does it mean that the function will be one to one ?

**thereddevils** · March 4th 2009, 05:42 AM

Originally Posted by HallsofIvy

(1) You know that sin x is periodic with period $2\pi$ . That should give you an easy answer.
(2) This is a little harder because this is one to one. I would show that by arguing that the derivative is always positive and so the function is increasing but I don't know what methods you have available.
(3) I don't recognise "In(x)". Do you mean the natural logarighm? That is ln(x)- "logarithm" starts with a "l" not "I"! If so, same comments as in (2) apply.

For (2) , i made a typo , Grandad has corrected it .

For (3) , yes it is a natural log . sorry for that mistake , ln

Thanks for pointing out .

**Grandad** · March 4th 2009, 05:43 AM

Hello thereddevils

Originally Posted by thereddevils

Thanks once again . For (2) If the function is one to one , does it mean that the function will be one to one ?

Sorry, I don't understand!

Grandad

**thereddevils** · March 4th 2009, 06:08 AM

Originally Posted by Grandad

Hello thereddevilsSorry, I don't understand!

Grandad

Sorry . If the function is an increasing function , does it mean that the function will be one to one ?

**Grandad** · March 4th 2009, 11:48 AM

Hello thereddevils

Originally Posted by thereddevils

Sorry . If the function is an increasing function , does it mean that the function will be one to one ?

Yes, providing it's a strictly increasing function - in other words, it doesn't have any sections where the gradient is zero. This is true of the function $g(x)$ .

Grandad

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