Find (f*g)(1) and (g*f)(1)?

(fg)(x) = (gf)(x) = (-2x-1)(x^2+2) = -2x^3 -4x - x^2 - 2

(fg)(1) = (gf)(1) = -2(1)^3-4(1)-(1)^2-2

= -2-4-1-2

= -9

For (f*g)(1) you first put 1 into the g function to get

1^2 + 2 = 3

THis result is then put into the f function

2*3 - 1 = 6 -1 = 5

For (g*f)(1) you first put 1 into the f function

2*1 - 1 = 2-1 = 1

this result is then put into the g function

1^2 + 2 = 3

To find (f*g)(1), you do f(g(1)). So, you do f(x^2 + 2). Then, to find f(x^2 + 2), you plug in the stuff in between the parenthesis into the x of the f(x) function. So, you'll get f(x^2 + 2) = -2(x^2 + 2) -1.

Therefore, (f*g)(1) = f(g(1)) = -2(x^2 + 2) -1.

To find (g*f)(1), you do g(f(1)). So, you do g(-2x + 1). Then, to find g(-2x + 1), you plug in the stuff between the parenthesis into the x of the g(x) function. So, you'll get g(-2x + 1) = (-2x + 1)^2 + 2.

Therefore, (g*f)(1) = g(f(1)) = (-2x + 1)^2 + 2

I'm assuming you mean composition, not multiplication

(f*g) = -2(x^2+2)-1 = -2x^2 - 4 - 1 = -2x^2 - 5

(g*f) = (-2x-1) ^2+2 = (4x^2 - 4x + 1) + 2 = 4x^2 - 4x + 3

now just plug the number 1 in and you got the answer

(f*g) = f[g(x)]

= f(x^2+2)

= 2(x^2+2) - 1

= 2x^2 +3

Therefore fg(1) = 2(1)(1)+3 = 5

(g*f) = g[f(x)]

= g(2x-1)

= (2x-1)^2+2

= 4x^2 -4x +3

Therefore gf(1) = 4(1)(1) - 4(1) +3 = 3

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RE:

Find (f*g)(1) and (g*f)(1)?

f(x)=-2x-1

g(x)x^2+2

(f*g)(1)

let * to be o

f o g(1)

f(g(1))

g(1) = 1 + 2

= 3

f(3) = -2*3 - 1

= -7