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Author: nsharp1
No. Questions: 3
Created: May 27, 2020
Last Modified: 2 years ago

Growth of Exponential Functions

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Proposition A:

If [math]f[/math] is an exponential function of the form [math]f(x) = a b^x[/math], [math]a>0, \ b>0 " and " b!=1[/math], and [math]\alpha > 0[/math] is a given constant, then [math](f(x+alpha))/f(x) = c[/math] for all values of [math]x in RR[/math], where [math]c[/math] is a real-valued constant.
Grade 11 Functions and Relations CCSS: HSF-LE.A.1, HSF-LE.A.1a
A.
Using the definition of [math]f(x)[/math] in the proposition, rewrite [math]f(x + \alpha) / f(x)[/math], in simplest terms.
  1. [math]2ab^{x+\alpha}[/math]
  2. [math]b^{x+2 \alpha}[/math]
  3. [math]b^alpha[/math]
  4. [math]\alpha[/math]
Grade 11 Functions and Relations CCSS: HSF-LE.A.1, HSF-LE.A.1a
B.
Which of the following reasons best explains why proposition A is true?
  1. Since [math]\alpha[/math] is a constant value, the value of [math]f(x+\alpha)/f(x)[/math] will also be a constant value.
  2. Because the simplified form of [math]f(x+\alpha)/f(x)[/math] is an exponential equation, it is valid for all values of [math]x in RR[/math], and will be constant.
  3. Since the simplified form of [math]f(x+\alpha)/f(x)[/math] is independent of [math]x[/math], it will be a constant value.
  4. Because a constant value to the power of a constant value is also a constant, the value [math]f(x+\alpha)/f(x)[/math] will also be a constant value.
Grade 11 Functions and Relations CCSS: HSF-LE.A.1, HSF-LE.A.1a
C.
One way to restate Proposition A is as follows: "Exponential functions grow by equal factors over equal intervals." What is the value of the equal factors and length of the equal intervals as represented in Proposition A?
  1. Equal factors are [math]x_0[/math], equal intervals are [math]c[/math].
  2. Equal factors are [math]alpha[/math], equal intervals are [math]x[/math].
  3. Equal factors are [math]alpha[/math], equal intervals are [math]c[/math].
  4. Equal factors are [math]c[/math], equal intervals are [math]alpha[/math].