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anonymous
 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^33X^212X7=k has exactly one real solution. Thanksss
anonymous
 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^33X^212X7=k has exactly one real solution. Thanksss

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0isnt it finding a value for k????

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol...tnx so for any real number \(k\) there is exactly one solution for the equation see the intersects of horizental lines and \(f(x)\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why do u differentiate the equation ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1345805312784:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because i want to see what is the behaviour of function

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if it was something like thisdw:1345805467267:dw for some values between \(k_1\) and \(k_2\) there is more than one real root

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so i differentiate it to see if there is some critical points like last drawing or not.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so is the answer k=all real numbers?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think yes...if im not wrong

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oops : misread : \[f(x)=2x^3−3x^2−12x−7\]\[f'(x)=6x^26x12=0\]it gives \(x^2x2=0\) so \(x=1,2\) so we have two critical points

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry @Snowflake123 ...

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1you sub the x back into the y for k i think..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so whats the k values

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1why not you try and sub it back in

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1sub it into y = 2x^2  3x^2  12x  7

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the graph will be like thisdw:1345806482815:dw just try to work it out

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the equation \(f'(x)\) has 2 roots so there are 2 k's we must work it out

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1according to my notes dont you find the roots from the stationary points? so there is only one line \(y=k\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for \(k=k_1,k_2\) 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 \(f(x)=k\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the region we lookin for is \[k>k_2 \ \text{&} \ k<k_1 \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what do u mean by '' there is only one line y=k''

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1woops forget it i made a mistake; youre right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol...i hope im right...

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1yes you are!! i just drew abit it differently so got kinda confused
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