# FNGR 2 | Lesson 5 | Try This (Inverse Relationships) # Inverse Relationships

In Exploring Function Reflections, when two functions are reflected over the line $$y=x$$ , then the point $$(a,b)$$ from one function and the point $$(b,a)$$ from the second function form a line segment perpendicular to the line $$y=x$$ and the points are equidistant to the line $$y=x$$.

If both curves are functions (they pass the vertical line test), then this relationship is defined as the inverse of two functions and they are reflected over the line $$y=x$$ which is the identity function.  (Note: identity means that for any input $$x$$, the output is $$x$$ or $$f(x)=x$$)

For example, if the point $$(1,2)$$ lies on the function $$f(x)$$ , then $$(2,1)$$ lies on the inverse of $$f(x)$$ .  We write the inverse function as  $$f^{-1}(x)$$ .  Note that the inverse of a function is not its reciprocal; therefore, $$f^{-1}(x) \neq \dfrac{1}{f(x)}$$.

Also, if a function has an inverse, then it is said to be 1-to-1.  A function $$f$$ is 1-to-1 if no two elements in the domain of $$f$$ correspond to the same element in the range of $$f$$ . In other words, each $$x$$ in the domain has exactly one image in the range.

Finally, a function that passes the horizontal line test (when horizontal lines do not intersect the graph of the function $$f$$ in more than one point), then the function is 1-to-1.

Determine which of the following are inverse functions and explain your reasoning. Graphs are created using Desmos.com.

1.  $$f(x) = -2x~–~5$$ and $$g(x) = -\dfrac x2 ~-~ \dfrac 52$$ 2.  $$h(x) = x + 5$$  and $$k(x)=-x-5$$ 3.  $$r(x)=x^3$$ and $$t(x)=\sqrt{x}$$ ##### Check your solutions here. 