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Here is the function: $$f(x) = \left\{ \begin{array}{ll} x^2 &\mbox{for } x \leq 0 \\ ax + b &\mbox{for } x > 0 \end{array} \right. $$

The question is, the function above has an inverse function for which of the following values of a and b?

a) $a = -1$, $b = -2$

b) $a = -1$, $b = 2$

c) $a = 0$, $b = -1$

d) $a = 1$, $b = -2$

e) $a = 1$, $b = 2$

I need to be able to solve this quickly for a test I'm going to take which will require solving questions like this in an average time of 90 seconds. Currently I'm using the method of plugging in the options and testing out different solutions and it takes me a lot longer than that to solve. Is there any information I should know about that reveals answers to questions like this faster?

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What characteristic must a function have to be a function? –  Shahar Jun 16 at 20:59
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3 Answers 3

You could graph it. If the graph fails the horizontal line test, the inverse will fail the vertical line test and so won't be a function.

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For something like this, I would just make sure the function is one-to-one / injective. This is because if a function is injective, then it has an inverse as well.

For this problem in particular, the range for $x^2$ takes up all the positive numbers so the range of the second function, $ax+b$ will have to be all negative numbers in order for $f(x)$ to preserve its injective property. We note that $ax+b$ is an injective function if $a \neq 0$.

For test taking methods, this would immediately mean $b$ would have to be negative so we can eliminate choice b) and e). $a$ would also have to be negative so choice c) and d) are eliminated.

Therefore we only have choice a) left. Notice how this was found through elimination from knowing what kind of answer I was looking for. I'm not a fan of plugging in answers as I'm sure test writers know most folks will get the correct answer through brute force, time be damned.

Edit: Mis-read the domain of each functions. Previously said choice d) was the correct answer.

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You know that the function will be 1-1 if its graph satisfies the horizontal line test, as mentioned in the previous answer, and you know that the graph of f for $x\le0$ is the left half of the parabola $y=x^2$, which opens up and has vertex at the origin.

Therefore the line $y=ax+b$ can only have negative values for $x>0$, which implies that the slope $a$ and the y-intercept $b$ must both be negative.

(Thus a) is the correct answer.)

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