Let f(x) = mx+b. Also, let g(x) be the inverse of f(x)
To find the inverse, we start with y = mx+b and swap x and y. From there, we solve for y like so
y = mx+b
x = my + b
x-b = my
my = x-b
y = (x-b)/m ... note m is in the denominator, so m cannot be 0
y = (x/m) - (b/m)
y = (1/m)x - (b/m)
g(x) = (1/m)x - (b/m)
This new equation is linear because it is in the form (slope)x+(y intercept)
The new slope is 1/m and the new y intercept is -b/m
So this proves that the inverse g(x) is linear when f(x) is linear.
The inverse function is:
[tex]y = \frac{x}{m} - \frac{b}{m}[/tex]
Since [tex]m \neq 0[/tex] and [tex]x \neq b[/tex], it is a linear function with slope [tex]\frac{1}{m}[/tex] and intercept [tex]-\frac{b}{m}[/tex].
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[tex]y = mx + b[/tex]
In which:
[tex]x = my + b[/tex]
[tex]my = x - b[/tex]
[tex]y = \frac{x}{m} - \frac{b}{m}[/tex]
Since [tex]m \neq 0[/tex] and [tex]x \neq b[/tex], it is a linear function with slope [tex]\frac{1}{m}[/tex] and intercept [tex]-\frac{b}{m}[/tex].
A similar problem is given at https://brainly.com/question/18594541