^{2}+ 48x - 10x - 30)

^{2}+ 38x -30)

f(x) = 8x - 5, g(x) = 2x+6

Find (fg)(x).

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Given;

f(x) = 8x - 5

g(x) = 2x+6

(fg)(x) = f(x) * g(x)

(fg)(x) = (8x - 5)(2x + 6)

(fg)(x) = (16x^{2} + 48x - 10x - 30)

(fg)(x) = (16x^{2} + 38x -30)

Hi Sue;

f(x) = 8x - 5, g(x) = 2x+6

Find (fg)(x)

Find (fg)(x)

(fg)(x)=(8x-5)(2x+6)

Let's FOIL...

FIRST...(8x)(2x)=16x^{2}

OUTER...(8x)(6)=48x

INNER...(-5)(2x)=-10x

LAST...(-5)(6)=-30

(fg)(x)=16x^{2}+48x-10x-30

This can also be written as...

Shelly J. | Excellent Maths Tutoring for academic successExcellent Maths Tutoring for academic su...

Hi Sue,

f(x)=8x-5

g(x)=2x+6

(fg)(x)=f(x)g(x)

=(8x-5)(2x+6)

=16x²+48x-10x-30

=16x²+38x-30

The answer by Parviz is incorrect for the problem as written.

(fg)(x) uses the same assumed notation as arithmetic. Therefore, in the same spirit as my previous response,

(fg)(x) = (f*g)(x) = f(x)*g(x) = (8x-5)(2x+6).

The problem that Parviz solved was f(g(x)) which is often written (f º g)(x). Notice, in particular the open circle that separates the functions f and g. It is shorthand for " of " and thus represents putting g(x) in where ever you would normally have placed "x" in the function f. This is the procedure that Parviz demonstrated, but was not the one that was given in the request.

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

f(x) = 8x -5 g(x) = 2x +6

f( g(x) )= 8( 2X +5) - 5

= 16X +35

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