154692

1 We need to find the range of values for which
2 f(x)=sqrt(x)+1/sqrt(x)<4
3 First, we find x values for which f(x)=4
4 sqrt(x)+1/sqrt(x)=4
5 (x+1)/sqrt(x)=4
6 x+1=4sqrt(x)
7 (x^2)+2x+1=16x
8 (x^2)-14x+1=0
9 By the quadratic formula, we have
10 x=(14(+or-)sqrt(196-4)))/(2)
11 Which simplifies to
12 x=7(+or-)4sqrt(3)
13 Since x>0 , x=7+4sqrt(3)
14 Now, we must analyse the derivative at this root
15 g(x)=(x^2)-14x+1
16 g'(x)=2x-14
17 g'(7+4sqrt(3))=2(7+4sqrt(3))-14
18 g'(7+4sqrt(3))=8sqrt(3)
19 The gradient at this root is positive, meaning that for x>7+4sqrt(3), g(x)>0
20 We want g(x)<0, so x<7+4sqrt(3)
21 But we also want x to be positive, therefore
22 0x<7+4sqrt(3)