- anonymous

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

- chestercat

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- anonymous

nice drawing :P

- anonymous

isnt it finding a value for k????

- anonymous

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

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## More answers

- anonymous

why do u differentiate the equation ?

- anonymous

|dw:1345805312784:dw|

- anonymous

because i want to see what is the behaviour of function

- anonymous

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

- anonymous

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

- anonymous

so is the answer k=all real numbers?

- anonymous

i think yes...if im not wrong

- anonymous

oops : misread :
\[f(x)=2x^3âˆ’3x^2âˆ’12xâˆ’7\]\[f'(x)=6x^2-6x-12=0\]it gives \(x^2-x-2=0\) so \(x=-1,2\) so we have two critical points

- anonymous

sorry @Snowflake123 ...

- Mimi_x3

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

- anonymous

exactly

- anonymous

so whats the k values

- Mimi_x3

why not you try and sub it back in

- Mimi_x3

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

- Mimi_x3

2x^3**

- anonymous

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

- Mimi_x3

why are there two k's?

- anonymous

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

- Mimi_x3

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

- anonymous

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

- anonymous

so the region we lookin for is \[k>k_2 \ \text{&} \ k

- anonymous

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

- Mimi_x3

woops forget it i made a mistake; youre right

- anonymous

lol...i hope im right...

- Mimi_x3

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

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