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This printable supports Common Core Mathematics Standard HSF-BF.B.4, HSF-BF.B.4b

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Verifying Inverse Functions By Composition (Grades 11-12)

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Verifying Inverse Functions By Composition

1. 
For [math]f(x) = 5x - 8[/math] and [math]g(x) = 1/5 x+ 8/5[/math], is [math]g(x)[/math] the inverse of [math]f(x)?[/math] Use composition of functions to verify.
  1. Yes
  2. No
2. 
Using composition of functions, is [math]g(x) = 2x-10[/math] the inverse function of [math]f(x) = 1/2x + 10 ?[/math]
  1. Yes
  2. No
3. 
For [math]f(x) = (x-4)/(6x)[/math] and [math]g(x) = 4/(6x+1)[/math], is [math]g(x)[/math] the inverse of [math]f(x) ?[/math] Use composition of functions to verify.
  1. Yes
  2. No
4. 
Using composition of functions, is [math]f(x) = x/(7+2x)[/math] the inverse of [math]g(x) = (-7x)/(2x-1) ?[/math]
  1. Yes
  2. No
5. 
If [math]f(x) = 3e^(x+2)[/math] and [math]g(x) = ln(x/3)-2[/math], is [math]g(x)[/math] the inverse of [math]f(x) ?[/math] Verify using composition of functions.
  1. Yes
  2. No
6. 
Using composition of functions, is [math]f(x) = 5root[3](8x-1)[/math] the inverse of [math]x^3/125 + 1/8[/math].
  1. Yes
  2. No
7. 
Let [math]f(x) = x^2[/math] and [math]g(x) = sqrt(x)[/math].
A. 
Find [math]h(x) = f(g(x))[/math] and state its domain.
  1. [math]h(x) = x, \ \ x >= 0[/math]
  2. [math]h(x) = |x|, \ \ RR[/math]
  3. [math]h(x) = x, \ \ RR[/math]
  4. [math]h(x) = |x|, \ \ x<=0[/math]
B. 
Find [math]h_2(x) = g(f(x))[/math] and state its domain.
  1. [math]h_2(x) = x, \ \ x>=0[/math]
  2. [math]h_2(x) = |x|, \ \ RR[/math]
  3. [math]h_2(x) = x, \ \ RR[/math]
  4. [math]h_2(x) = |x|, \ \ x<=0[/math]
C. 
Is [math]g(x)[/math] the inverse of [math]f(x)?[/math] Explain.
  1. Yes, since both [math]h(x)[/math] and [math]h_2(x)[/math] are equal to [math]x[/math].
  2. No, because the domain of [math]h(x)[/math] and [math]h_2(x)[/math] are different.
  3. No, because [math]h(x) != x[/math].
  4. No, because [math]h_2(x) != x[/math].
D. 
When finding [math]h(x)[/math], what is the importance of the domain?
  1. It has no special importance.
  2. It shows that, even though [math]h(x)[/math] and [math]h_2(x)[/math] seem to be the same function, they have different domains.
  3. It is necessary to see that the domain is all real numbers because the domain of inverse functions, when composed with the original function, needs to be all real numbers.
  4. It means that the absolute value sign can be dropped, since the domain is zero and positive numbers, so the function simply becomes [math]x[/math].

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