1. ## Inverse function values

Hello again!

I've been posting a lot of questions recently and I just wanted to let you know that I truly appreciate it! I'm doing some power studying and am forgetting all of these things now that it's summer. Grarrrhh... I got another one for you...

$\displaystyle f(x) = \begin{cases} x^2 ----- for x \leq 0 \\ ax + b --- for x > 0\end{cases}$

The function f above has an inverse function for which of the following values of a and b?
Answer: a= -1, b = -2

Now, I know how to find the inverse of a function. Basically, you flip the x and y values and then try and put it back into the y= form. But I'm lost about the whole values thing... could somebody explain to me how they arrived at that answer?

2. Are these multiple choice questions? I can see that there are many values of and b for which the inverse exists....

Give us the remaining choices and we will tell you why those will not have inverses...

P.S: What is "power" studying?

3. "Power" studying is cramming info before a test.

Other options:

(A) the one I just posted
(B) a= -1, b = 2
(C) a = 0, b = -1
(D) a = 1, b = -2
(E) a = 1, b = 2

Thanks again!

4. Originally Posted by theimcq
"Power" studying is cramming info before a test.

Other options:

(A) the one I just posted
(B) a= -1, b = 2
(C) a = 0, b = -1
(D) a = 1, b = -2
(E) a = 1, b = 2

Thanks again!
For a function to have inverse, it should be both one-one and onto.
I think graphing is the best solution. Try graphing (A) and you will see that its both one-one and onto. So end of story...

(B) not one-one
(C) not onto
(D) not one-one and not onto
(E) not one-one and not onto

Graph the functions:
The function is 1-1 if every horizontal line cuts the graph at most once.
The functions is onto if every horizontal line cuts the graph at least once.

5. Thanks for the very helpful post. However, on this test they ("the proctors") won't provide a graphing calculator (it's actually disallowed). Because of this, I would need to know how to solve it algebraically? How do you solve this algebraically?

6. Originally Posted by theimcq
Thanks for the very helpful post. However, on this test they ("the proctors") won't provide a graphing calculator (it's actually disallowed). Because of this, I would need to know how to solve it algebraically? How do you solve this algebraically?
I didnt ask you to use a graphing calculator. The equations are equations of a parabola and a straight line. Dont you know to plot these things ?

7. To answer this problem you need to recognize that y=ax+b defines a line with slope a and y-intercept at b. For the inverse of a function to be a function there must be a unique value of x associated with each y. Given that $\displaystyle y=x^2$ for x<0 covers the range $\displaystyle 0<=y < \infty$, that means that the y=ax+b part must be limited to 0 and negative numbers. Therore a must be negative, and b must be 0 or negative. Hence the answer.