Question

Help with finding values of function

Answer 1

@aum

Answer 2

combine the two rational expressions on the right into one by fining the LCD.

Answer 3

Least common denominator?

Answer 4

yes.

Answer 5

a/b + c/d = (ad + bc) / bd

Answer 6

I know how to find the common how would you find the least?

Answer 7

doesn't matter. common denominator will do.

Answer 8

oks it would be multiplying the first by 5 and the second by 3

Answer 9

\[
f(x) = \frac{2}{x-3} + \frac{1}{x-5} = \frac{2(x-5)+1(x-3)}{(x-3)(x-5)} =
\frac{2x-10+x-3}{(x-3)(x-5)} = \frac{3x-13}{(x-3)(x-5)}
\]
f(x) is not defined when x = 3 and x = 5.

f(x) = 0 when the numerator is zero. That is, 3x - 13 = 0, x = 13/3.

Answer 10

the end of the equation went off the screen

Answer 11

But it is all visible. Nothing past that.

Answer 12

\(\Large x = \frac{13}{3} = 4\frac 13 \)

Answer 13

so you combined the equations and simplified. but i am confused with the last part. f(x) = 0

Answer 14

i now you have to use f(x)= 0 for the problem but what about after that

Answer 15

\[f(x) = \frac{3x-13}{(x-3)(x-5)}\]
f(x) = 0 when \[\frac{3x-13}{(x-3)(x-5)} = 0; ~~~~~ \text{when }3x - 13 = 0; ~~~~x = 13/3\]

Answer 16

so you set it equal to 0 and then just solved the equation. that makes sense but what about finding the values o.o

Answer 17

There is just one value of x that will make f(x) = 0 and that x is 13/3.

Answer 18

So since it only requires it for 0 the only value would be 13/3?

Answer 19

They have given you a function f(x) and asked you to solve the equation f(x) = 0. We did that and found x = 13/3. That is it.

It is like asking what would make f(x) = x^2 - 5x + 6 = 0. So we solve the quadratic equation and find x = 3 and x = 2 will make f(x) zero. Those are the roots. Those are the values where a plot of f(x) will cross the x-axis (where y = 0).

Same thing. If the original f(x) is plotted, it will cross the x-axis at just one point: x = 13/3. There are no other roots for this function.

Answer 20

Ok. Thank you for taking the time to explain this to me

Answer 21

You are welcome.