Verifying Inverse Functions By Composition
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Verifying Inverse Functions By Composition Answer Key
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.
- Yes
- No
2.
Using composition of functions, is [math]g(x) = 2x-10[/math] the inverse function of [math]f(x) = 1/2x + 10 ?[/math]
- Yes
- 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.
- Yes
- No
4.
Using composition of functions, is [math]f(x) = x/(7+2x)[/math] the inverse of [math]g(x) = (-7x)/(2x-1) ?[/math]
- Yes
- 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.
- Yes
- 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].
- Yes
- 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.
- [math]h(x) = x, \ \ x >= 0[/math]
- [math]h(x) = |x|, \ \ RR[/math]
- [math]h(x) = x, \ \ RR[/math]
- [math]h(x) = |x|, \ \ x<=0[/math]
B.
Find [math]h_2(x) = g(f(x))[/math] and state its domain.
- [math]h_2(x) = x, \ \ x>=0[/math]
- [math]h_2(x) = |x|, \ \ RR[/math]
- [math]h_2(x) = x, \ \ RR[/math]
- [math]h_2(x) = |x|, \ \ x<=0[/math]
C.
Is [math]g(x)[/math] the inverse of [math]f(x)?[/math] Explain.
- Yes, since both [math]h(x)[/math] and [math]h_2(x)[/math] are equal to [math]x[/math].
- No, because the domain of [math]h(x)[/math] and [math]h_2(x)[/math] are different.
- No, because [math]h(x) != x[/math].
- No, because [math]h_2(x) != x[/math].
D.
When finding [math]h(x)[/math], what is the importance of the domain?
- It has no special importance.
- 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.
- 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.
- 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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