# Why is my answer wrong? (Inverse Functions)

Find the inverse function of $g(x)=-4x+1$.

So I replace $g(x)$ with $y$, then solve for $x$:

$$4x=1-y\\ x = \frac{1-y}4\\ y = \frac{1-x}4$$ The answer was $g(x)^{-1} =(-1/4)x + 1/4$.

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$$1-x/4\neq (1-x)/4.$$ You forgot to divide everything by 4, as far as I can tell from your notation. – Ian Coley Mar 13 '14 at 1:43
That was a typo. It was suppose to be (1-x)/4. – igknighton Mar 13 '14 at 1:45
Your method and answer are essentially correct. I would skip the switcheroo between $x$ and $y$, though. If $y=g(x)$, then you should write the inverse function as $x=g^{-1}(y)$. Note also you should correct the placement of the "$-1$"--it should be written as $g^{-1}(y)$, not $g(y)^{-1}$. The latter signifies $\frac{1}{g(y)}$. Otherwise good! – MPW Mar 13 '14 at 2:00

$$g(x)=-4x+1\\ \mbox{Let g(x)=y. Then, }-4x=y-1\\ x=(y-1)/(-4)\\ x=\dfrac{1-y}{4}\\ \boxed{g^{-1}(x)=\dfrac{1-x}{4}}$$ Or, simplifying, $$g^{-1}(x)=\dfrac{1-x}{4}\\ g^{-1}(x)=\dfrac{1}{4}-\dfrac{x}{4}=\dfrac{1}{4}-\dfrac{1}{4}x\\ \boxed{g^{-1}(x)=-\dfrac{1}{4}x+\dfrac{1}{4}}$$