Answer:
2.
Step-by-step explanation:
1. Recall the definition of one-to-one function. We say that [tex]f[/tex] is one-to-one if and only if the equality [tex]f(a)=f(b)[/tex] implies that [tex]a=b[/tex].
First example: [tex] f(x)=\frac{x-1}{3x+3}[/tex].
We start with
[tex] f(a) = \frac{a-1}{3a+3} = \frac{b-1}{3b+3} = f(b)[/tex].
In the above equalities, the central one, making cross products, is equivalent to
[tex] (a-1)(3b+3) = (3a+3)(b-1)[/tex].
Then, [tex]3ab+3a-3b-3 = 3ab-3a+3b-3[/tex]. Now, cancelling identical terms we get [tex] 3a-3b=-3a+3b[/tex] which is equivalent to [tex]6a=6b[/tex]. Thus, simplifying the factor 6, we get [tex]a=b[/tex]. Therefore, the function is one-to-one.
Second example: [tex] f(x) = \sqrt{5x+9}[/tex].
We try to follow the same reasoning. We start with
[tex]\sqrt{5a+9} = \sqrt{5b+9}[/tex].
Now, we elevate both members to the square (recall that both expressions are positive, so there is no problem with squaring) and obtain:
[tex] 5a+9=5b+9[/tex]. We eliminate the 9's and we get [tex] 5a=5b[/tex]. Simplifying the 5's, we finally obtain that [tex] a=b[/tex]. Therefore the function is one-to-one.
Third example. [tex]f(x) = \frac{7}{4x^2}[/tex].
This example is different from the previous ones. Recall that we prove that a function is not one-to-one if we find two different numbers [tex]a\neq b[/tex] such that [tex]f(a)=f(b)[/tex].
Then, notice that if we take [tex]a=-1[/tex] and [tex]b=1[/tex] we get that
[tex]f(-1) = \frac{7}{4(-1)^2} = \frac{7}{4}[/tex] and [tex]f(-1) = \frac{7}{4\cdot 1^2} = \frac{7}{4}[/tex].
Then, this function is not one-to-one.
Fourth example: [tex]f(x) = \frac{1}{2}x^3[/tex].
As the first two cases. Consider [tex]\frac{1}{2}a^3=\frac{1}{2}b^3[/tex]. Then, simplify the factor 1/2, and we get [tex]a^3=b^3[/tex].
Now, let us transform the previous equality into [tex]a^3-b^3=0[/tex]. This expression can be factored: [tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]. Thus,
[tex]0=(a-b)(a^2+ab+b^2)[/tex]
Notice that the above equality is possible if and only if [tex]a-b=0[/tex] or
[tex]a^2+ab+b^2=0[/tex]. But the second possibility is impossible (you can check it using the general formula for second degree equations), thus, necessarily [tex]a-b=0[/tex] which is equivalent to [tex]a=b[/tex]. Therefore, the function is one-to-one.
Fifth example: [tex]f(x) = 3x^4+7x^3[/tex].
First, notice that the polynomial [tex]3x^4+7x^3[/tex] can be factored as
[tex]3x^4+7x^3 = x^3(3x+7)[/tex].
From here is not difficult to deduce that [tex] f(0)=0=f(-7/3)[/tex] and consequently the function is not one-to-one.
2. As we are looking for the inverse function [tex]f^{-1}[/tex] of [tex]f(x)[/tex] is a better idea to use the following notation:
[tex]y = \sqrt[3]{x-2} + 8[/tex],
instead of [tex]f(x)[/tex].
Notice that, with this notation we are saying that [tex]y[/tex] depends on [tex]x[/tex]. Now, when we are looking for the inverse function we will try to write the variable [tex]x[/tex] in terms of [tex]y[/tex]. With this idea, the first step is to move the 8 in the right hand side to the left hand side, so:
[tex]y-8 = \sqrt[3]{x-2}[/tex].
Recall that we need to isolated the variable [tex]x[/tex], so we must ‘‘eliminate’’ the cubic root. In order to do this we take cubic powers in both side of the above equality. Then,
[tex](y-8)^3 = x-2[/tex].
Finally, we take the 2 to the left hand side and we get
[tex]x = (y-8)^3+2[/tex].
Thus, the inverse function is [tex]f^{-1}(x) = (x-8)^3+2[/tex].
Notice that if we do [tex]f(f^{-1}(x))[/tex] we get [tex]x[/tex].
