Inequality and max question (1 Viewer)

[Seraph]

Why is the inequality x^2 - 6 _< |X|

= -3 _< x _< 3

wouldnt it be

-2 _< x _< 3



also how would i differentiate and find the stationary pt for this equation y= 1/2x(25-10x)^(1/2)
this is what i did
y ' = EVENTUALLY will equal

- 5 / (2(25-10x)^(1/2) + (25-10x)^(1/2) / 2

which equals 10x + 50 - 20x / 4(25-10x)^(1/2)
which i thought equals 10x + 50 - 20x = 0
and x = 5
indeed i was wrong....

 

[acmilan]

Im doing it in my head so it might be wrong. Is the stationary point x = 10/3?

 

[Xayma]

Why is the inequality x^2 - 6 _< |X|

= -3 _< x _< 3

wouldnt it be

-2 _< x _< 3

x²-6 &le; |x|
x²-6 &le; x OR 6-x² &ge; x
x²-x-6 &le; 0 OR x²+x-6 &ge; 0
(x-3)(x+2) &le; 0 OR (x+3)(x-2) &ge; 0

It is true between x=2 and x=3 AND x=-2 and x=-3 AND x=2 and x=-2.

Therefore

-3&le;x&le;3

 

[~*HSC 4 life*~]

also how would i differentiate and find the stationary pt for this equation y= 1/2x(25-10x)^(1/2)
...

is that meant to be 1/2x... as in 'x' or multiply?

 

[acmilan]

and if it is means "x" does it mean 1 over 2x or a half x

 

[Seraph]

hmmm (x/2)(25-10x)^(1/2)


thanks Xayma

 

[acmilan]

Ok then ignore my previous answer

 

[flower_farie]

do u know the answer for that? i got some absurd answer involvin the quadratic formula to work out the x values...i got x=5/2...dont think thats right tho

 

[Seraph]

yes that is right

so you ended up usign the quadratic formula??

what a bitch of a question can someone show me the working for this ... ive never really had problems with differentiation like this before

 

[flower_farie]

were u talking about the '97 hsc paper 6a?...the answer was supposed to be 5/3 *my bad* when u differentiate it using the produst rule, u bring the -5x/2(25-20x)^1/2 over so it becomes a postive...then cross multiply...u didnt need to use the quadratic formula