Question

What’s the inverse function

Answer 1

Answer:

$f^{-1}(x)=-2x+6$.

Step-by-step explanation:

$y=f(x)$

$y=3-\frac{1}{2}x$

The biggest thing about finding the inverse is swapping x and y.  The inverse comes from switching all the points on the graph of the original.  So a point (x,y) on the original becomes (y,x) on the original's inverse.

Sway x and y in:

$y=3-\frac{1}{2}x$

$x=3-\frac{1}{2}y$

Now we want to remake y the subject (that is solve for y):

Subtract 3 on both sides:

$x-3=-\frac{1}{2}y$

Multiply both sides by -2:

$-2(x-3)=y$

We could leave as this or we could distribute:

$-2x+6=y$

The inverse equations is $y=-2x+6$.

Now some people rename this $f^{-1}$ or just call it another name like $g$.

$f^{-1}(x)=-2x+6$.

Let's verify this is the inverse.

If they are inverses then you will have that:

$f(f^{-1}(x))=x \text{ and } f^{-1}(f(x))=x$

Let's try the first:

$f(f^{-1}(x))$

$f(-2x+6)$  (Replace inverse f with -2x+6 since we had $f^{-1})(x)=-2x+6$

$3-\frac{1}{2}(-2x+6)$  (Replace old output, x, in f with new input, -2x+6)

$3+x-3$  (I distributed)

$x$

Bingo!

Let's try the other way.

$f^{-1}(f(x))$

$f^{-1}(3-\frac{1}{2}x)$ (Replace f(x) with 3-(1/2)x since $f(x)=3-\frac{1}{2}x$)

$-2(3-\frac{1}{2}x)+6$ (Replace old input, x, in -2x+6 with 3-(1/2)x since $f(x)=3-\frac{1}{2}x$)

$-6+x+6$ (I distributed)

$x$

So both ways we got x.

We have confirmed what we found is the inverse of the original function.

Answer 2

Answer:

$\laege\boxed{f^{-1}(x)=-2x+6}$

Step-by-step explanation:

$f(x)=3-\dfrac{1}{2}x\to y=3-\dfrac{1}{2}x\\\\\text{Exchange x to y and vice versa:}\\\\x=3-\dfrac{1}{2}y\\\\\text{Solve for}\ y:\\\\3-\dfrac{1}{2}y=x\qquad\text{subtract 3 from both sides}\\\\-\dfrac{1}{2}y=x-3\qquad\text{multiply both sides by (-2)}\\\\\left(-2\!\!\!\!\diagup^1\right)\cdot\left(-\dfrac{1}{2\!\!\!\!\diagup_1}y\right)=-2x-3(-2)\\\\y=-2x+6$