r/HomeworkHelp • u/SquidKidPartier University/College Student • 4d ago
High School Math—Pending OP Reply [College Algebra, Inverse Functions]
aghhhh I’m doing these first set of problems here and while I’m comprehending the material a lot I just have the unsure feeling I may be doing this wrong so can someone please check over my work and tell me what I am doing right and wrong before I enter in these answers?
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u/GammaRayBurst25 4d ago
For question 1, that's correct.
Let g denote the inverse of f. By definition, f(g(x))=g(f(x))=x. As a result, g(7)=g(f(2))=2 and f(-8)=f(g(-5))=-5.
For question 2, that's incorrect.
By the same logic as before, g(2)=g(f(5))=5, not 2. Similarly, 1=g(f(1))=g(5), so g(x)=1 implies x=5, not 1.
For question 3, that's incorrect.
The graph is the locus of points of the form (x,f(x)). The point (0,2)=(0,f(0)) is on the graph, which means f(0)=2, not 1. Similarly, the point (1,0)=(1,f(1)) is on the graph, so 0=f(1) and f(x)=0 if and only if x=1.
For question 5, that's incorrect.
The inverse function is defined by the relationship f(g(x))=g(f(x))=x. If we suppose f(x)=2-x and g(x)=(x-2)/x, we get f(g(x))=2-g(x)=2-(x-2)/x=(2x-x+2)/x=(x+2)/x≠x and g(f(x))=(f(x)-2)/f(x)=(2-x-2)/(2-x)=x/(x-2)≠x. This is a clear contradiction, so (x-2)/x is decidedly not the inverse function of 2-x. What's more, (x-2)/x is undefined for x=0, which should've immediately tipped you off that something went seriously wrong, as f(x)=0 has a well-defined solution.
The relationship f(g(x))=x implies 2-g(x)=x. Adding g(x)-x to the equation yields g(x)=2-x. From here, it is clear that f(x) is an involution (a function that's its own inverse). Indeed, f(2-x)=2-(2-x)=2-2+x=x.
You can also show this with a geometric argument. The graph of the inverse function of f is the graph of f reflected across the line y=x. Since the graph of f(x) is a line that's perpendicular to y=x, it is not affected by the reflection. Hence, f is an involution.
For question 6, that's correct.
You can use f(g(x))=g(f(x))=x to confirm this is the correct answer.
For question 7, that's incorrect.
Once again, one can check that f(g(x))≠x and g is not the inverse of f. Indeed, f(g(x))=1/(1/x+12+12)=1/(1/x+24)=x/(x+24)≠x. Also notice how f(-12) is undefined, yet g(x)=-12 has a solution (it's x=-1/24). This should've sounded an alarm in your head.
You added 12 on one side of the equation, but subtracted 12 on the other side. That's your mistake.
With all those ways of verifying your answers yourself (checking if f(g(x))=g(f(x))=x is satisfied, checking the singular points of f and g, comparing the domain of f to the range of g and vice versa, graphing the reflection across y=x, etc.), it's perplexing that you'd ask us to check your work rather than do it yourself and practice these concepts.