Asked by Ashley
Given that the equation x^2(3-x) has 3 real solutions of k, give the set of possible values for k.
Answers
Answered by
Steve
no equation
no k
no way to help
no k
no way to help
Answered by
Ashley
x^2(3-x)=k. That's the equation.
Answered by
Steve
Well, we know that
y=x^2(3-x)=0 has roots at 0,0,3
And, we know that the graph just touches the x-axis at (0,0) the double root, so cannot have a small k<0 because that will have only a single root near x=3.
So, x^2(3-x) = k only has 3 roots if k>=0
Now, if you have calculus as a tool, you can see that y' = 3x(2-x) so there is a maximum at (2,4). So, we must have k<=4.
So, y=x^2(3-k) has 3 roots only if 0<=k<=4
y=x^2(3-x)=0 has roots at 0,0,3
And, we know that the graph just touches the x-axis at (0,0) the double root, so cannot have a small k<0 because that will have only a single root near x=3.
So, x^2(3-x) = k only has 3 roots if k>=0
Now, if you have calculus as a tool, you can see that y' = 3x(2-x) so there is a maximum at (2,4). So, we must have k<=4.
So, y=x^2(3-k) has 3 roots only if 0<=k<=4
Answered by
Ashley
Ok, thank you.
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