How to do these type of questions without graphs

Question

The function $f:[0,3]\to[1,29]$ is defined by $$f(x)=2x^3-15x^2+36x+1$$

Then the function is

$1.$ one one and onto

$2.$ onto but not one one

$3.$ one one but not onto

$4.$ neither one one nor onto

The way I know to approach these type of questions is using a rough hand drawn graph. But I can't graph this function by hand.

Is there any way to do these types of questions in general without a graph$?$ This is just one question, I'm facing problems in many questions like these$?$

Please tell me the method if there's any. Any help is greatly appreciated.

you really can draw this by hand, and the practice is good for you. First find the values at the integers involved, $x=0,1,2,3$ — Will Jagy, Jun 08 '23 at 17:53

Another User · Accepted Answer · 2023-06-08T17:46:06.143

3

Since $$ f'(x)=6x^2-30x+36=6(x-2)(x-3), $$ you know that $f'(x)<0$ when $x\in(2,3)$ and that $f'(x)>0$ when $x\in[0,2)$. Therefore, $f$ is strictly increasing on $[0,2]$ and strictly decreasing on $[2,3]$. So, $\max f=f(2)=29$. Besides, since $f(0)=1$ and $f(3)=28$, the range of $f$ is $[1,29]$. Therefore, the correct choice is the second one.

edited Jun 08 '23 at 17:46

answered Jun 08 '23 at 17:39

Another User

5,048

The correct answer is onto but not one one as given in the solution – MathStackexchangeIsNotSoBad Jun 08 '23 at 17:45
That is correct. I had read $[0,29]$ instead of $[1,29]$. – Another User Jun 08 '23 at 17:46
Can you please add some more details like why you did this and how is the increasing or decreasing nature deciding whether it's one one or not or onto or not? – MathStackexchangeIsNotSoBad Jun 08 '23 at 17:47
I understood why it's not one one..pls explain for onto – MathStackexchangeIsNotSoBad Jun 08 '23 at 17:48
Since $f(0)=1$ and $f(2)=29$, it follows from the intermediate value theorem the every number from $[1,29]$ must belong to the range of $f$. – Another User Jun 08 '23 at 17:49
I haven't learnt ivt yet..but by logic i think I'm understanding..but pls add some details – MathStackexchangeIsNotSoBad Jun 08 '23 at 17:50
Thanks...I understood by logic – MathStackexchangeIsNotSoBad Jun 08 '23 at 17:50

How to do these type of questions without graphs

1 Answers1