{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

bmhicks_homework_2

bmhicks_homework_2 - Brandy Hicks MTHSC 974.001 Homework 2...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Brandy Hicks August 31, 2007 MTHSC 974.001 - Homework # 2 1. For r > o , draw the graph of f 1 r ( X ) = 1 2 rx 2 , | x | ≤ 1 r | x | - 1 2 r , | x | ≥ 1 r to see if f 1 r is convex on R . 1.JPG Although the drawing is rudimentary, it appears that f 1 r ( x ) is convex on R (a) What happens as r + ? Proof: As r → ∞ , f 1 r ( x ) → -∞ . As r gets very large, 1 r gets very close to zero. Since the value of f 1 r ( x ) = 1 2 rx 2 when | x | ≤ 1 r , then when the absolute value of x is between 0 and 1 r , which is very close to 0, we get 1 2 · r (which remember is approaching ) · x 2 (and x is very close to 0). This means that the equation basically becomes 1 2 · ∞ · 1 × 10 - 100000000000 . Therefore this whole piece of the function is going to 0. Now we look at the the other segment of the function. We see that when | x | ≥ 1 r , | x | - 1 2 · r . So | x | has to be greater than 1 r which is basically 1 , which is very close to zero. This then makes our function | x | (which is greater than 0) - 1 2 ·∞ .
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}