## Snowflake123 3 years ago Not sure how to do this: Please explain??? :D Write down the set of values of the constant k for which the equation 2X^3-3X^2-12X-7=k has exactly one real solution. Thanksss

1. SNSDYoona

nice drawing :P

2. Snowflake123

isnt it finding a value for k????

3. mukushla

lol...tnx so for any real number $$k$$ there is exactly one solution for the equation see the intersects of horizental lines and $$f(x)$$

4. SNSDYoona

why do u differentiate the equation ?

5. mukushla

|dw:1345805312784:dw|

6. mukushla

because i want to see what is the behaviour of function

7. mukushla

if it was something like this|dw:1345805467267:dw| for some values between $$k_1$$ and $$k_2$$ there is more than one real root

8. mukushla

so i differentiate it to see if there is some critical points like last drawing or not.

9. Snowflake123

so is the answer k=all real numbers?

10. mukushla

i think yes...if im not wrong

11. mukushla

oops : misread : $f(x)=2x^3−3x^2−12x−7$$f'(x)=6x^2-6x-12=0$it gives $$x^2-x-2=0$$ so $$x=-1,2$$ so we have two critical points

12. mukushla

sorry @Snowflake123 ...

13. Mimi_x3

you sub the x back into the y for k i think..

14. mukushla

exactly

15. Snowflake123

so whats the k values

16. Mimi_x3

why not you try and sub it back in

17. Mimi_x3

sub it into y = 2x^2 - 3x^2 - 12x - 7

18. Mimi_x3

2x^3**

19. mukushla

so the graph will be like this|dw:1345806482815:dw| just try to work it out

20. Mimi_x3

why are there two k's?

21. mukushla

the equation $$f'(x)$$ has 2 roots so there are 2 k's we must work it out

22. Mimi_x3

according to my notes dont you find the roots from the stationary points? so there is only one line $$y=k$$

23. mukushla

for $$k=k_1,k_2$$ there are exactly 2 real roots for the equation $$f(x)=k$$ for $$k_1<k<k_2$$ there are exactly 3 real roots for the equation $$f(x)=k$$

24. mukushla

so the region we lookin for is $k>k_2 \ \text{&} \ k<k_1$

25. mukushla

what do u mean by '' there is only one line y=k''

26. Mimi_x3

woops forget it i made a mistake; youre right

27. mukushla

lol...i hope im right...

28. Mimi_x3

yes you are!! i just drew abit it differently so got kinda confused