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Snowflake123
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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
 one year ago
 one year ago
Snowflake123 Group Title
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
 one year ago
 one year ago

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SNSDYoona Group TitleBest ResponseYou've already chosen the best response.0
nice drawing :P
 one year ago

Snowflake123 Group TitleBest ResponseYou've already chosen the best response.0
isnt it finding a value for k????
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.4
lol...tnx so for any real number \(k\) there is exactly one solution for the equation see the intersects of horizental lines and \(f(x)\)
 one year ago

SNSDYoona Group TitleBest ResponseYou've already chosen the best response.0
why do u differentiate the equation ?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.4
dw:1345805312784:dw
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.4
because i want to see what is the behaviour of function
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.4
if it was something like thisdw:1345805467267:dw for some values between \(k_1\) and \(k_2\) there is more than one real root
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.4
so i differentiate it to see if there is some critical points like last drawing or not.
 one year ago

Snowflake123 Group TitleBest ResponseYou've already chosen the best response.0
so is the answer k=all real numbers?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.4
i think yes...if im not wrong
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.4
oops : 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
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.4
sorry @Snowflake123 ...
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
you sub the x back into the y for k i think..
 one year ago

Snowflake123 Group TitleBest ResponseYou've already chosen the best response.0
so whats the k values
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
why not you try and sub it back in
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
sub it into y = 2x^2  3x^2  12x  7
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.4
so the graph will be like thisdw:1345806482815:dw just try to work it out
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
why are there two k's?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.4
the equation \(f'(x)\) has 2 roots so there are 2 k's we must work it out
 one year ago

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

mukushla Group TitleBest ResponseYou've already chosen the best response.4
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 \(f(x)=k\)
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.4
so the region we lookin for is \[k>k_2 \ \text{&} \ k<k_1 \]
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.4
what do u mean by '' there is only one line y=k''
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
woops forget it i made a mistake; youre right
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.4
lol...i hope im right...
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
yes you are!! i just drew abit it differently so got kinda confused
 one year ago
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