If f(x)=1/9x-2, what is f-1(x)

1 answer:

3 0

To find the inverse of f(x), you solve for the independent variable, x, and then swap the variable labels. The reason for doing so is easily seen if one remembers what an inverse function really is. The inverse function will produce the opposite coordinate as the parent function. If f(x)=(x,y), f^-1(x)=(y,x), so the inverse function simply creates the inverse of all of the points that exist for f(x)...with that out of the way, let y=f(x), then:

y=1/(9x-2) (I am trying to decipher what you actually meant to type)

Multiply both sides by (9x-2)

y(9x-2)=1 divide both sides by y

9x-2=1/y add 2 to both sides

9x=1/y+2 which is equal to

9x=(1+2y)/y divide both sides by 9

x=(1+2y)/(9y) now switch labels...

y=(1+2x)/(9x)

f^-1(x)=(1+2x)/(9x)

Again, I am not sure that I understood what you meant to type for the equation... strictly as typed you actually have:

y=x/9-2 which can be written as:

y=(x-18)/9 multiply both sides by 9

9y=x-18 add 18 to both sides

9y+18=x then switch labels

y=9x+18

f^-1(x)=9x+18

The denominator must be made clear in this format by using () brackets, otherwise it is anyone's guess...

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Find the values of x for which the combined function h(x) = is undefined if f(x) = 5x + 1 and g(x) = x2 - 9.

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We need to know the definition of the "combined function" h(x). I'm going to guess--by looking at answers--that the function is $h(x)=\frac{f(x)}{g(x)}$ making it

$h(x)=\frac{5x+1}{x^2-9}$

This function is undefined wherever the denominator is equal to 0 (division by zero is undefined). Factor the denominator.

$h(x)=\frac{5x+1}{(x+3)(x-3)}$

The two values of x that make the denominator 0 are 3 and -3, otherwise written $x=\pm 3$ .