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1.) Let \(f\) be defined by

 

                                    \(f(x) = \left\{ \begin{array}{cl} 3-x & \text{ if } x \leq 3, \\ -x^3+2x^2+3x & \text{ if } x>3. \end{array} \right.\)

 

Calculate \(f^{-1}(0)+f^{-1}(6)\)

 

 

2.) If \(f(x)=\frac{x+4}{x^2+ax+b}\) and \(f(x)\) has two vertical asymptotes at x=1 and x=-2, then find the sum of a and b.

 Aug 21, 2020
 #1
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0

1. f^{-1}(0) + f^{-1}(6) = 2 + 1 = 3.

 

2. a + b = (4) + (-2) = 2.

 Aug 21, 2020
 #2
avatar+109 
-1

I appreciate that you've attempted to answer this question, however, both of your answers are wrong. I feel like you just picked some random numbers and showed some useless steps. But, this doesn't work. Sorry!!

surpriseindecisionsadangelangry

dp1806  Aug 21, 2020
 #5
avatar+1035 
+2

@geust, I'm not sure if you understand, but f^-1 means the inverse of the function, not f to the power of -1

 

:)

ilorty  Aug 23, 2020
 #3
avatar+30947 
+1

This graph should help with question 1:

 Aug 21, 2020
 #6
avatar+1035 
+2

If you can't see the picture, this link should work...https://web2.0calc.com/img/upload/86da579d92c1f599a70c4342/g0.jpg

im not sure why the image is blocked...

:)

ilorty  Aug 23, 2020
 #4
avatar+30947 
+1

For question 2, the vertical asymptotes will occur where the denominator is zero.  These will be where the roots of the denominator are. i.e. they will be such that (x + 2)(x - 1) = 0

 

Expand this and compare with x^2 + ax + b =0,  to find a and b.

 Aug 21, 2020

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