𝑓(𝑥)=𝑥+4f(x)=x+4𝑔(𝑥)=3𝑥2−7g(x)=3x 2 −7Find (𝑓⋅𝑔)(𝑥)(f⋅g)(x).A.(𝑓⋅𝑔)(𝑥)=3𝑥3+12𝑥2−7𝑥−28(f⋅g)(x)=3x 3 +12x 2 −7x−28B.(𝑓⋅𝑔)(𝑥)=3𝑥3+28(f⋅g)(x)=3x 3 +28C.(𝑓⋅𝑔)(𝑥)=3𝑥3−28(f⋅g)(x)=3x 3 −28D.(𝑓⋅𝑔)(𝑥)=3𝑥3+12𝑥2+7𝑥+28(f⋅g)(x)=3x 3 +12x 2 +7x+28SUBMITarrow_backPREVIOUS

𝑓(𝑥)=2𝑥2+4𝑥−5f(x)=2x 2 +4x−5𝑔(𝑥)=6𝑥3−2𝑥2+3g(x)=6x 3 −2x 2 +3Find (𝑓−𝑔)(𝑥)(f−g)(x).A.(𝑓−𝑔)(𝑥)=6𝑥3−4𝑥2−4𝑥+8(f−g)(x)=6x 3 −4x 2 −4x+8B.(𝑓−𝑔)(𝑥)=−6𝑥3+4𝑥2+4𝑥−8(f−g)(x)=−6x 3 +4x 2 +4x−8C.(𝑓−𝑔)(𝑥)=6𝑥3+4𝑥−2(f−g)(x)=6x 3 +4x−2D.(𝑓−𝑔)(𝑥)=−6𝑥3+4𝑥2+4𝑥−2(f−g)(x)=−6x 3 +4x 2 +4x−2SUBMITarrow_backPREVIOUS

𝑓(𝑥)=2𝑥3+3𝑥2−7𝑥+2f(x)=2x 3 +3x 2 −7x+2𝑔(𝑥)=2𝑥−5g(x)=2x−5Find (𝑓+𝑔)(𝑥)(f+g)(x).A.(𝑓+𝑔)(𝑥)=2𝑥3+3𝑥2+5𝑥+3(f+g)(x)=2x 3 +3x 2 +5x+3B.(𝑓+𝑔)(𝑥)=2𝑥3+3𝑥2−5𝑥−3(f+g)(x)=2x 3 +3x 2 −5x−3C.(𝑓+𝑔)(𝑥)=2𝑥3+3𝑥2−5𝑥+3(f+g)(x)=2x 3 +3x 2 −5x+3D.(𝑓+𝑔)(𝑥)=2𝑥3+3𝑥2+5𝑥−3(f+g)(x)=2x 3 +3x 2 +5x−3SUBMITarrow_backPREVIOUS

𝑓(𝑥)=2𝑥3+5𝑥2−6𝑥−7f(x)=2x 3 +5x 2 −6x−7𝑔(𝑥)=2𝑥−3g(x)=2x−3Find (𝑓−𝑔)(𝑥)(f−g)(x).A.(𝑓−𝑔)(𝑥)=2𝑥3+5𝑥2−8𝑥+4(f−g)(x)=2x 3 +5x 2 −8x+4B.(𝑓−𝑔)(𝑥)=2𝑥3+5𝑥2−4𝑥+4(f−g)(x)=2x 3 +5x 2 −4x+4C.(𝑓−𝑔)(𝑥)=2𝑥3+5𝑥2−8𝑥−4(f−g)(x)=2x 3 +5x 2 −8x−4D.(𝑓−𝑔)(𝑥)=2𝑥3+5𝑥2−4𝑥−4(f−g)(x)=2x 3 +5x 2 −4x−4SUBMITarrow_backPREVIOUS

𝑓(𝑥)=∣2𝑥+3∣−5f(x)=∣2x+3∣−5𝑔(𝑥)=7g(x)=7Find (𝑓−𝑔)(𝑥)(f−g)(x).A.(𝑓−𝑔)(𝑥)=∣2𝑥+3∣−12(f−g)(x)=∣2x+3∣−12B.(𝑓−𝑔)(𝑥)=∣2𝑥∣−9(f−g)(x)=∣2x∣−9C.(𝑓−𝑔)(𝑥)=∣2𝑥−9∣(f−g)(x)=∣2x−9∣D.(𝑓−𝑔)(𝑥)=∣2𝑥−2∣−7(f−g)(x)=∣2x−2∣−7SUBMITarrow_backPREVIOUS

Find the value of:(𝑓𝑔)(2)−(𝑔𝑓(4))(fg)(2)−( fg​ (4))left parenthesis, f, g, right parenthesis, left parenthesis, 2, right parenthesis, minus, left parenthesis, start fraction, g, divided by, f, end fraction, left parenthesis, 4, right parenthesis, right parenthesisGiven the functions:𝑓(𝑥)=1𝑥𝑔(𝑥)=3𝑥+1f(x)g(x)​ = x1​ =3x+1​