# Spivak Calculus Chapter 7 problem 1(v)

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1. Jul 13, 2016

### Derek Hart

1. The problem statement, all variables and given/known data
Decide whether the given function is bounded above or below on the given interval, and which take on their maximum or minimum value. (Notice that ƒ might have these properties even if ƒ is not continuous, and even if the interval** isn't closed)

**The interval is (-a-1, a+1), and it is assumed in the problem that a>-1 so that -a-1 < a+1.

2. Relevant equations
ƒ(x) = x^2 if x< or = a, a+2 if x>a

3. The attempt at a solution
I concluded that the function is in all cases of a>-1 is bounded above and below, and that if a< or =-1/2, then a+2 is the maximum and minimum, or if -1/2< a < or = 0, then a^2 was the minimum, and then that if a> or = 0 then 0 is the minimum.
The part where I struggle is with the latter 2 maximum values where, in the answer book, it states that "Since a+2 > (a+1)^2 only when [(-1-√5)/2] < a < [(-1+√5)/2], when a> or = -1/2 the function ƒ has a maximum value only for a< or = [(-1+√5)/2] (the maximum value being a+2)". I simply don't understand how he came up with the first inequality for a. I tried rearranging the inequality between a+2 and (a+1)^2 but i can't figure it out.

2. Jul 13, 2016

### andrewkirk

Start with $a+2>(a+1)^2$. Expand the square and move everything onto the RHS of the inequality, so that you'll get

0>[a quadratic formula in $a$]

The RHS has a positive coefficient for $a^2$, so it will be positive everywhere except between the two roots of the quadratic equation you get when you set the RHS equal to zero.
Those two roots are the numbers Spivak uses in his inequality above.

3. Jul 14, 2016

### Derek Hart

Ahh thanks, I feel stupid now. I haven't encountered the quadratic formula in any of the problems i have completed and haven't used it in a long time so it slipped my mind.