Question

Suppose that the functions f and g are defined as follows:

See pic
Find f-g and g(x) . Then, give their domains using interval notation.

Answer 1

@Luigi0210 can you show me how to do this?

Answer 2

O is (2x)° R is (3y+8) and Q is (2x+4)

Answer 3

@AwesomeAries @Luigi0210

Answer 4

I don't understand what you mean or how you got that @adrianaR can you explain it to me

Answer 5

@Auxuris can you help me with this

Answer 6

\(\bf f(x)=5x^2-4\qquad g(x)=\cfrac{1}{\sqrt{3x-1}}\\ \quad \\ \quad \\

(f-g)(x)\implies f(x)-g(x)\implies (5x^2-4)-\left(\cfrac{1}{\sqrt{3x-1}}\right)\\ \quad \\

(f\cdot g)(x)\implies f(x)\cdot g(x)\implies (5x^2-4)\cdot \left(\cfrac{1}{\sqrt{3x-1}}\right)\)

keep in mind that for a fraction, when the denominator is 0, the fraction is "undefined"
so that places a restriction on the INPUT "x"
"x" cannot be a value that makes the denominator 0

and for a square root, when the radicand is negative, the root is "imaginary"
so that places another restriction on INPUT "x"
"x" cannot be a value that makes the radicand negative

Answer 7

so once you get (f-g)(x) and (f*g)(x) keep in mind that the INPUT or domain is anything but values that makes the denominator 0 or the radicand negative

Answer 8

so, (f-g)(x)=\[\frac{ 15x^3-5x^2+4-12x-\sqrt{\left( 3x-1 \right)} }{ 3x-1 }\] domain is \[\left( \frac{ 1 }{ 3 }, \infty \right)\]
(f*g)(x)=\[\frac{ 5x^2-4 }{ \sqrt{3x-1} }\] domain:(\[\left( \frac{ 1 }{ 3 }, infty \right)\])

Answer 9

is that right?

Answer 10

@ganeshie8

Answer 11

(f*g)(x) is correct

Answer 12

domain for both is correct

Answer 13

but for \(f-g\), dont simplify the function.
enter it as-it-is ok ?

Answer 14

nvm, ur work looks more good :)

Answer 15

excellent job !! enter ur work itself !!!!