Stuck on Differential equation?

Question

Im stuck on this differentiation equation, Ill post what I have done, im close to the answer in the back, but not there.

The question is Given that for f(x) = (x^5/2)/5 + 64/x, find the values of x and y when f'(x) = 0

well the f'(x) means dy/dx which of the above is

1/5 x^5/2 + 64x^-1

dy/dx = 5/10 x^3/2 -64x^-2 = f'(x)

Now if the f'(x) = 0 then 5/10 x^3/2 - 64x^-2 = 0

Now im not factorising so I solved for x and did

1/2 x ^3/2 = 64/x^2

(1/2)(x^3/2)(x^2) = 64
(x^3/2)(x^2) = 64/(1/2)
x^7/2 = 128
x^7 = 128^2
x^7 = 16384
x = 16384^1/7
x = 4

Now much to my horror the answer in the back is 4.57

No idea where I have gone wrong....any pointers would be great

Thanks for all your help.

Just for info the answer in the back says x=4.57 y = 19.59

The second part of the question says: Find the value of d2y/dx^2 at the point where f'(x) = 0.

So you differentiate again to get 3/4 x^1/2 + 128x^-3.

My answer is 3.5, the answer in the back of the boox is 2.34 which in no way corresponds to the above...do the rest of you agree with this?

Substituting x = 4 gives

Answer 1

No, there is nothing wrong with your working x = 4 is fine.

Answers in the back are sometimes wrong!

Have you double checked the question?

Answer 2

y = f(x)=(x^5/2)/5 + 64/x
f'(x) = (1/2)x^3/2-64/x^2

When f'(x) = 0,

(1/2)x^3/2 = 64/x^2

So x^7/2 = 128 = 2^7
So x^1/2 = 2
So x=4.

So y = (4^5/2)/5+(64/4)= 22.4

It looks like you are right, and the book is wrong.

Answer 3

f´(x) =1/2 x^3/2-64/x^2=0 so
1/2x^7/2-64=0 x^7/2= 128 =2^7 so x= (2^7)^2/7= 4
The back is wrong

Answer 4

Your solution is good. You can prove it by substituting 4.000 into the original equation. You get y=22.400. This is the minimum because if you substitute 4.1 or 3.9 you get a y>22.400. 4.57 is clearly wrong.

Answer 5

4y'' = -4(k^2)cos(kt) -25y = -25cos(kt) So 4y'' = -25y implies -4(k^2) = -25. So k^2 = 25/4, giving you k = +/- 5/2.

Answer 6

I think the book is wrong... I see no mistakes in any of your work.