Spivak Calculus Chapter 7 problem 1(v)

In summary, the given function, ƒ(x) = x^2 if x≤a, a+2 if x>a, is bounded above and below for all values of a>-1. If a≤-1/2, then a+2 is both the maximum and minimum value. For -1/2<a≤0, a^2 is the minimum value. For a>0, 0 is the minimum value. The maximum value is only defined for a≤(√5-1)/2, where a+2 is the maximum value. This is found by using the quadratic formula to solve for the roots of the inequality a+2>(a+1)^2.
  • #1
Derek Hart
14
1

Homework Statement


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.

Homework Equations


ƒ(x) = x^2 if x< or = a, a+2 if x>a

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.
 
Physics news on Phys.org
  • #2
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.
 
  • Like
Likes Derek Hart
  • #3
andrewkirk said:
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.
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.
 

1. What is the main concept behind the Spivak Calculus Chapter 7 problem 1(v)?

The main concept behind this problem is to find the derivative of a given function using the definition of derivative.

2. What is the definition of derivative used in this problem?

The definition of derivative used in this problem is the limit definition, which is the slope of a tangent line to a curve at a given point.

3. How do you approach solving this problem?

To solve this problem, you need to first identify the given function and use the definition of derivative to find the limit as h approaches 0. Then, simplify the expression and substitute the given value to find the derivative.

4. What are the common mistakes made in solving this problem?

Some common mistakes made in solving this problem include not identifying the given function correctly, not understanding the definition of derivative, and making errors in simplifying the expression or substituting the value.

5. How can this problem help improve understanding of derivatives?

This problem can help improve understanding of derivatives by reinforcing the fundamental concept of using the definition of derivative to find the slope of a tangent line. It also helps to develop problem-solving skills and identify common mistakes that can be avoided in future problems.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
478
  • Calculus and Beyond Homework Help
Replies
5
Views
991
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
698
  • Calculus and Beyond Homework Help
Replies
2
Views
466
  • Calculus and Beyond Homework Help
Replies
2
Views
549
Replies
9
Views
721
  • Calculus and Beyond Homework Help
Replies
32
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
320
Back
Top