Composition and Inverse
MAT222
Composition and Inverse
The following functions will be used in this week’s assignment. 1. fx=2x+5 2. gx=x2-3 3. hx=7-x3
We are first asked to compute f-h4. Like many algebra problems, the solution becomes obvious, or at least easier, once it has been rewritten into a different format. We will solve the problem in this way: f-h(4) - Original expression f4-h4 - Rewritten
24+5-7-43 - Substitute 4 for x
8+5- 33 - Simplify
13-1 - Simplify further
12 -Solution The second problem involves composing two pairs into each other. f°g(x) - Original expression f(gx) - Rewritten so that rule of f will apply fx2-3 - g inserted
2x2-3+5 - f applied
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To solve h-1(x): hx=7-x3 - Original function y=7-x3 - Rewritten with y replacing function name x=7-y3 - Rewritten with x and y transposed
3x=7-y - Multiply both sides by 3
-y=3x-7 - Isolate y y=-3x+7 - multiply both sides by -1 h-1x=-3x+ 7 - Solved. After reading chapter 11, the real-world functions that pertain to the force on a ship's sails, e.g. the location of the center of effort (propulsive force), the direction of that force and the pressure on those sails. One of my hobbies is researching naval history, specifically in the late 18th and early 19th century (I was exposed to C.S. Forester at an early age). However, outside of a few dinghys and sailboats, I only have an amateur's understanding of the forces involved in directing a ship of any size with and against the wind. It surprised me to find a great deal of physics and algebra is involved in determining the maximum force that a sail can handle. I was aware that trigonometry was routinely used in determining the ship's position, and that advanced mathematics were necessary to use a sextant to determine longitude. Further math was needed to estimate position based on earlier known position, and the ship speed versus leeway, resulting in dead reckoning. However, the functions involved in the breaking point and most effective points of sail were unknown to me before reading this chapter.
References
Ashford
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