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Let \[f(x) = \begin{cases} x^2+9 &\text{if }x<-5, \\ 3x-8&\text{if }x\ge-5. \end{cases} \]If $f(x)=10$, find the sum of all possible values of $x$.

 Aug 29, 2017
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\( \[f(x) = \begin{cases} x^2+9 &\text{if }x<-5, \\ 3x-8&\text{if }x\ge-5. \end{cases} \]If $f(x)=10$\)

 

Find the sum of all possible values of x

 

If   x^2 +  9  = 10

Then x^2  = 1

And x  =  ± 1     .....   but for this function,  x < -5,  so neither of these are good

 

If  3x  - 8  = 10

3x  = 18

x  = 6  .......   and  x  ≥ -5

 

So....the sum of all possible values of x  =  6

 

 

 

cool cool cool

 Aug 29, 2017

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