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name: nc.m4.af.a.1, nc.m4.af.a.2 date: unit 2 • piecewise functions, co…

Question

name: nc.m4.af.a.1, nc.m4.af.a.2
date:
unit 2 • piecewise functions, composition of functions, and regression
lesson 2.1: piecewise, step, and absolute value functions
practice 2.1: piecewise, step, and absolute value functions
the following graph is called a phase diagram. it shows the pressure and temperature at which a substance is a gas, a liquid, a solid, or all three phases of matter. the pressure scale, p(t), is in thousands, and the temperature scale t is in degrees celsius. use the graph to complete problems 1–4.
graph of phase diagram with pressure (in thousands) on y - axis and temperature (°c) on x - axis, with points a, b, c, d, e, f and regions for solid, liquid, gas

  1. why is the whole graph not a combination or piecewise function?
  2. which points could be used to define individual functions from the data on the graph? determine the coordinates of those points.
  3. what are the coordinates of the point on the graph at which all three phases of the substance exist?
  4. what function type(s) can be used to model these data points?

Explanation:

Brief Explanations
  1. A function requires each input (temperature $T$) to have exactly one output (pressure $P(T)$). For many $T$ values in the phase diagram, there are multiple valid $P(T)$ values (e.g., at $T=80^\circ\text{C}$, pressure can be below, at, or above 225 thousand, corresponding to gas, solid/liquid/gas, or liquid phases). This violates the definition of a function.
  2. Individual functions are defined by continuous, single-output segments. The segments are $A\to B\to C$, $C\to E\to F$, so their endpoints are the points that define each function segment. Coordinates are read from the graph axes (temperature on x-axis, pressure in thousands on y-axis).
  3. The point where all three phases coexist is the intersection of all three phase boundary lines, which is point $C$. Its coordinates are read from the graph.
  4. The segment $A\to B\to C$ has a constant positive slope, matching a linear function. The segment $C\to E\to F$ curves downward with a decreasing slope, matching an exponential (or inverse/non-linear decreasing) function.

Answer:

  1. It fails the vertical line test: some $T$ values map to multiple $P(T)$ values, violating the function definition.
  2. The points are $A$, $B$, $C$, $E$, $F$. Their coordinates are:
  • $A=(0, 100)$
  • $B=(40, 150)$
  • $C=(80, 225)$
  • $E=(60, 325)$
  • $F=(40, 550)$
  1. $(80, 225)$
  2. Linear function (for $A\to B\to C$) and exponential/non-linear decreasing function (for $C\to E\to F$)