We have to remember that in one-to-one functions each element of the domain maps to a different element in the range.
When we have a bijection we have that each element in the set A, for example, corresponds to exactly one element of B, and vice versa.
When we have a bijection, we also have an inverse function.
Then, we have that:
[tex]g(x)=-2x-13[/tex]
To find the inverse, we have to interchange the variables:
[tex]y=-2x-13[/tex]
Now, we have:
[tex]x=-2y-13[/tex]
And we need to solve for y:
1. Add 13 to both sides of the equation:
[tex]x+13=-2y-13+13\Rightarrow x+13=-2y[/tex]
2. Divide both sides by -2:
[tex]\frac{(x+13)}{-2}=\frac{-2y}{-2}\Rightarrow-\frac{(x+13)}{2}=y[/tex]
Then, we have that:
[tex]g^{-1}(x)=-\frac{(x+13)}{2}[/tex]
We can check this if we make a composition between the two functions (then we will get x as a result).
[tex](g^{-1}\circ g)=g^{-1}(g(x))=-\frac{((-2x-13)+13)_{}}{2}=-\frac{(-2x)}{2}=\frac{2x}{2}=x[/tex]
Then, we have that the result for this composition is equal to x. Thus:
[tex](g^{-1}\circ g)(-3)=-3^{}[/tex]
We also have that:
We have that h(4) = 3, then h^(-1)(3) = 4 or
[tex]h(4)=3\Rightarrow h^{-1}(3)=4[/tex]
In summary, we have:
[tex]g^{-1}(x)=-\frac{(x+13)}{2}[/tex][tex](g^{-1}\circ g)(-3)=-3[/tex][tex]h^{-1}(3)=4[/tex]
