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avatar+865 

Let

 

\(f(x) = \left\{ \begin{array}{cl} 2x + 1 & \text{if } x \le 3, \\ 8 - 4x & \text{if } x > 3. \end{array} \right.\)

 

Find the sum of all values of $x$ such that $f(x) = 0.$

I tried to do some guess and check (it worked on another problem i did similar to this) but it didn't really work...

 Jun 19, 2020
 #1
avatar+9673 
0

When x <= 3,

 

f(x) = 0

2x + 1 = 0

x = -1/2

 

When x > 3,

 

f(x) = 0

8 - 4x = 0

x = 2 

but x has to be greater than 3, so this is not actually a solution.

 

The sum of all values of x such that f(x) = 0 is -1/2.

 Jun 19, 2020
 #2
avatar+865 
0

Wow that was so fast thank you!

AnimalMaster  Jun 19, 2020
 #3
avatar+1262 
+4

so we use algebra that there is no soluton for if x>3 and thre is one solution for x<=3 which is -1/2 since we get 2x=-1 and the only way for x>3 is 2 but 2 is not greater than 3 so it is -1/2+[nothing]=-1/2

 Jun 20, 2020

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