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# 1.) Let be defined by

0
69
6
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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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1. f^{-1}(0) + f^{-1}(6) = 2 + 1 = 3.

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

Aug 21, 2020
#2
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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!!

dp1806  Aug 21, 2020
#5
+1035
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@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
+30947
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This graph should help with question 1:

Aug 21, 2020
#6
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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
+30947
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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