Answer:
[tex]\textsf{Inverse function}: \quad f^{-1}(x)=\dfrac{3}{4}-\dfrac{3}{10}x[/tex]
[tex]x\textsf{-intercept}:\quad\left(\dfrac{5}{2},0\right)[/tex]
Step-by-step explanation:
Given:
[tex]f(x)=2\frac{1}{2}-3\frac{1}{3}x[/tex]
Rewrite the function so it is a rational function
Convert the mixed numbers to improper fractions:
[tex]\implies f(x)=\dfrac{5}{2}-\dfrac{10x}{3}[/tex]
Make the denominators the same:
[tex]\implies f(x)=\dfrac{3 \cdot 5}{3\cdot 2}-\dfrac{2 \cdot 10x}{2 \cdot 3}[/tex]
[tex]\implies f(x)=\dfrac{15}{6}-\dfrac{20x}{6}[/tex]
Combine:
[tex]\implies f(x)=\dfrac{15-20x}{6}[/tex]
The inverse of a function is its reflection in the line y = x
To find the inverse, make x the subject
Replace f(x) with y:
[tex]\implies y=\dfrac{15-20x}{6}[/tex]
[tex]\implies 6y=15-20x[/tex]
[tex]\implies 6y-15=-20x[/tex]
[tex]\implies x=\dfrac{-6y+15}{20}[/tex]
[tex]\implies x=\dfrac{15-6y}{20}[/tex]
Replace x with [tex]f^{-1}(x)[/tex] and y with x:
[tex]\implies f^{-1}(x)=\dfrac{15-6x}{20}[/tex]
If necessary, convert back into the same format as the original function:
[tex]\implies f^{-1}(x)=\dfrac{15}{20}-\dfrac{6x}{20}[/tex]
[tex]\implies f^{-1}(x)=\dfrac{3}{4}-\dfrac{3}{10}x[/tex]
The x-intercept of the inverse function is the point at which it crosses the x-axis, so when [tex]f^{-1}(x)=0[/tex]
[tex]\implies \dfrac{15-6x}{20}=0[/tex]
[tex]\implies 15-6x=0[/tex]
[tex]\implies 6x=15[/tex]
[tex]\implies x=\dfrac{15}{6}=\dfrac{5}{2}[/tex]
Therefore, the x-intercept is:
[tex]\left(\dfrac{5}{2},0\right)[/tex]
