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Ok thanks
2. ### Maximum and minimum values

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.
3. ### Maximum and minimum values

Erm I have found my mistakes and able to show it but an additional question is can I use the sign test to find the maximum and minimum values from this kind of equation?sorry to bother you and thanks.
4. ### Maximum and minimum values

Yup both x values are yielding minima,if I don't use the calculator,it would be really difficult to sub it into the 2nd derivative. It has took me several hours but I am still unable to work it out.
5. ### Maximum and minimum values

I don't think I know how do I get x^3=8+-2(14)^{1/2} from the equations provided

Oooo
7. ### Maximum and minimum values

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
8. ### Maximum and minimum values

This is another method that I attempted to do but it is wrong
9. ### Maximum and minimum values

I just tried the 2nd derivative 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 (how I find y values,I get it from y=(x^2)/2,substituting the x values as stated above) into my second derivative but in the end I got all...
10. ### Maximum and minimum values

Yup I sub back into the original y equation and get another value of x and y ,however I just noticed that I don't actually need it right? Because I can do a 2nd derivative and put the previous x and y values into it and find whether it is maximum or minimum values,but again may I use a sign test...
11. ### Maximum and minimum values

Oo then by using the 3 values of x,I do a second derivative test and sub the values in order to prove that the x values are maximum and minimum values and we can show that maximum and minimum value of y occurs at those x right? And can I do a sign test to test whether y is decreasing or increasing?
12. ### Maximum and minimum values

I haven't attempt which I didn't thought of it that but as I sub it back into y^3, I get x=2 and y=2,so now I have 3 values of x and I can find 3 values of y,and then I need to sub into the second derivative test or using a sign test but If I do a sign test there,there are two variables,how to...
13. ### Maximum and minimum values

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,then I substitute x=2+2(14)^(1/6) and x=2-2(14)^(1/6) to get y values right? 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...

Yup correct
15. ### Maximum and minimum values

I think I have posted the attachment now,sorry I am a first timer here
16. ### Maximum and minimum values

Y^3=6xy-x^3-1
17. ### Maximum and minimum values

It should be +on the top and minus on the bottom,not sure how to write it and I have been using the sign test to test it,however the sign test should be for y values,but only x values are given by the question,plus do I sub the x values into the y^3 equation to find y or should I use dy/dx...
18. ### Maximum and minimum values

Homework Statement Show that the maximum and minimum values of y occurs when x^3=8+- 2(14)^1/2 Homework Equations The Attempt at a Solution