- #1
Liang Wei
- 18
- 0
Homework Statement
Show that the maximum and minimum values of y occurs when x^3=8+- 2(14)^1/2
Liang Wei said:I think I have posted the attachment now,sorry I am a first timer here
Right.Liang Wei said:As I know at stationary point,dy/dx=0 so from dy/dx=(2y-x^2)/(y^2-2x),from there I can find y=(x^2)/2
Ok that will fetch you the maximum and minimum of y.then I substitute x=2+2(14)^(1/2) and x=2-2(14)^(1/2) to get y values right?
I'm not sure if I understood this one.Then I reciprocate the dy/dx =dx/dy and then I do the sign test? But I got all + signs which means I fail to prove y has maximum and minimum values.
PhysicoRaj said:Right.
Ok that will fetch you the maximum and minimum of y.
I'm not sure if I understood this one.
But as soon as you get the ##y=x^2/2##, can you use it and the 'given' to solve for values of x?
You mean you substituted y=x^2/2 in the original y equation. What did it yield? Which are the three x's and y's?Liang Wei said:but as I sub it back into y^3
How did you get this? Probably by the previous x^3 equation which they have given?Liang Wei said:I have used the calculator to change the x^3 values into 15.84 and -5.84
but when I substitute the x values and y values into my second derivative but in the end I got all positive value which means only local minimum values only.
Liang Wei said:Yeah,the x values are from the x^3 that they provide,when you substitute those values,you will find that there are many zero values so it is able to get the d2y/dx2= +value which is greater than zero so both values of x are local minimums
PhysicoRaj said:I think they expect you to arrive at ##x^3=8+-2(14)^{1/2}## rather than assume it at the first. This is what is going wrong.
Liang Wei said:I don't think I know how do I get x^3=8+-2(14)^{1/2} from the equations provided
Liang Wei said:can I use the sign test to find the maximum and minimum values from this kind of equation?
Oh that one, for that you need to know the interval, and it requires you to solve for x.Liang Wei said:Sign test is a test which you use a value which lies inside the interval of a variable,example x then substitute it into dy/dx to find the sign and we can find the maxima and minima.
but anyways I manage to prove it by using 2nd derivative but not sure correct or wrong.
The maximum value of y is 12 when x is 8 + 2(14)^1/2.
The minimum value of y is -12 when x is 8 - 2(14)^1/2.
The maximum and minimum values of y can be calculated by plugging in the given value of x into the equation y = x^3 and solving for y.
Yes, the maximum and minimum values of y can be negative depending on the value of x. In this case, the maximum value of y is positive while the minimum value of y is negative.
No, there is no specific range for the values of x in this equation. As long as x is a real number, the equation will have a maximum and minimum value for y.