Snowflake123
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
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SNSDYoona
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nice drawing :P
Snowflake123
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isnt it finding a value for k????
mukushla
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lol...tnx
so for any real number \(k\) there is exactly one solution for the equation
see the intersects of horizental lines and \(f(x)\)
SNSDYoona
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why do u differentiate the equation ?
mukushla
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|dw:1345805312784:dw|
mukushla
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because i want to see what is the behaviour of function
mukushla
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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
mukushla
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so i differentiate it to see if there is some critical points like last drawing or not.
Snowflake123
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so is the answer k=all real numbers?
mukushla
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i think yes...if im not wrong
mukushla
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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
mukushla
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sorry @Snowflake123 ...
Mimi_x3
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you sub the x back into the y for k i think..
mukushla
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exactly
Snowflake123
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so whats the k values
Mimi_x3
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why not you try and sub it back in
Mimi_x3
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sub it into y = 2x^2 - 3x^2 - 12x - 7
Mimi_x3
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2x^3**
mukushla
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so the graph will be like this|dw:1345806482815:dw| just try to work it out
Mimi_x3
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why are there two k's?
mukushla
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the equation \(f'(x)\) has 2 roots so there are 2 k's we must work it out
Mimi_x3
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according to my notes dont you find the roots from the stationary points? so there is only one line \(y=k\)
mukushla
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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 \(f(x)=k\)
mukushla
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so the region we lookin for is \[k>k_2 \ \text{&} \ k<k_1 \]
mukushla
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what do u mean by '' there is only one line y=k''
Mimi_x3
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woops forget it i made a mistake; youre right
mukushla
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lol...i hope im right...
Mimi_x3
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yes you are!! i just drew abit it differently so got kinda confused