Question

prove that, for all x>1,

2*sqrt(x) >3-1/x @Mathematics

Answer 1

could you solve it using and a > b case, and a=b case and an a<b case?

Answer 2

spose a<b, but that a,b > 1

2sqrt(a) > 3 -1/b

4a > 9 -6/b -1/b^2

b^2 4a -9b^2+6b+1 > 0

b^2(4a -9) + 6b + 1 > 0

b = -6+- sqrt(36-4(1)(4a-9)) /2

b = -6+sqrt(36-16a +36)) /2

b = -6+2sqrt(18-4a) /2

b = -3+sqrt(18-4a)


i hate proofs, really i do .....

Answer 3

any particular strategy?

Answer 4

it wants us to use the first and second derivative to prove it using limits at infinity

Answer 5

let \[a = 2\sqrt {x}\]\[b = 3 - 1/x\]then let
\[y = a ^{2} = 4x\]\[z = b ^{2} = (3 - 1/x)(3 - 1/x) = (3x - 1)(3x - 1)/x ^{2} = (9x ^{2} - 6x + 1)/x ^{2}\]\[ = 9 - 6/x + 1/x ^{2}\]
when x = 1, y = 4 and z = 9 - 6 + 1 = 4
as x increases, z tends to 9 as both 1/x and 1/x^2 tend to zero
so, for x > 1, y > z

Answer 6

thanks for your help!

Answer 7

well, the first derivative would tell us how fast it changes and teh second tells us about the speed at which the speed is changeing

Answer 8

f(x) = 2sqrt(x); g(x)=3 -1/x

prove that f(x) > g(x), for all x>1

Answer 9

f'(x) = 1/sqrt(x) ; g'(x) = 1/x^2

f''(x) = -1/2sqrt(x^3) ; g''(x) = -2/x^3

if i did those right

Answer 10

infinity in the bottoms would both equate to 0 for both derivatives so i got no idea how that helps

Answer 11

lim(x->inf) f(x) = inf

lim(x->inf) g(x) = 3

that might be useful

Answer 12

thank you @amistre64!

Answer 13

youre welcome, good luck ;)