+0

# some help...

-1
91
3
+798

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
+8341
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
+798
0

Wow that was so fast thank you!

AnimalMaster  Jun 19, 2020
#3
+1128
+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