which pair of functions is not a pair of inverse functions?

a. f(x)= x+1/6 and g(x)= 6x-1

b. f(x)= x-4/19 and g(x)= 19x+4

c. f(x)= x5 and g(x)= 5√x

d. f(x)= x/x + 20 and g(x)= 20x/x-1
you!

Option in D is not an inverse function.

Step-by-step explanation:

A. Say,

⇒ 6y = x + 1

⇒ x = 6y - 1

Here, g(x) = 6x - 1, if g(x) is the inverse function of f(x).

B. Let us assume,

⇒ 19y = x - 4

⇒ x = 19x + 4

If g(x) is the inverse function of f(x), then, g(x) = 19x + 4

C. Let

⇒

Here, , if g(x) is the inverse function of f(x).

D. Let

⇒ yx + 20y = x

⇒ x(y - 1) = - 20y

⇒

Now, if g(x) is the inverse function of f(x), then,

Therefore, option in D is not an inverse function. (Answer)

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

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