1. ## Another function

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) $\displaystyle f(x)=sin x$ , x are real numbers

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

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

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

but for the rest , i am not really sure .

2. ## Range of function

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) $\displaystyle f(x)=sin x$ , x are real numbers

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

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

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

but for the rest , i am not really sure .
(1) $\displaystyle f$ is not one-to-one; e.g. $\displaystyle f(0) = f(\pi)$

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

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

3. 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) $\displaystyle f(x)=sin x$ , x are real numbers

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

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

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 $\displaystyle 2\pi$. 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.

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

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

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

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

5. Originally Posted by HallsofIvy
(1) You know that sin x is periodic with period $\displaystyle 2\pi$. 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 .

6. 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!

Hello thereddevilsSorry, I don't understand!

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 $\displaystyle g(x)$.