Jika f(2x+1/x-3) = x^2+2x-3 , maka nilai dari f'(0) adalah... a. -2 1/4 b. -2 c. -1 3/4 d. -1 1/4 e. -1

Mapel : Matematika
Kelas : XI SMA
Bab : Differensial

Pembahasan :
f(2x + 1 / x - 3) = x² + 2x - 3

Misalkan (2x + 1)/(x - 3) = a
x = (3a + 1)/(a - 2)

Maka...
f(a) = [(3a + 1)/(a - 2)]² + 2[(3a + 1)/(a - 2)] - 3
f(a) = (9a² + 6a + 1)/(a² - 4a + 4) + (6a + 2)/(a - 2) - 3
f(a) = (9a² + 6a + 1 + 6a² - 10a - 4 - 3a + 6)/(a - 2)
f(a) = (15a² - 7a + 3)/(a - 2)
f'(a) = 0
f'(0) = 0

[tex]\displaystyle \text{misal :}\\\frac{2x+1}{x-3}=a\\\frac{3a+1}{a-2}=x\\\\f\left(\frac{2x+1}{x-3}\right)=x^2+2x-3\\f\left(\frac{2x+1}{x-3}\right)=(x-1)(x+3)\\f(a)=\left(\frac{3a+1}{a-2}-1\right)\left(\frac{3a+1}{a-2}+3\right)\\f(a)=\left(\frac{3a+1-a+2}{a-2}\right)\left(\frac{3a+1+3a-6}{a-2}\right)\\f(a)=\left(\frac{2a+3}{a-2}\right)\left(\frac{6a-5}{a-2}\right)\\f(x)=\frac{(2x+3)(6x-5)}{(x-2)^2}\\f(x)=\frac{12x^2+8x-15}{x^2-4x+4}[/tex]

[tex]\displaystyle f'(x)=\frac{(24x+8)(x^2-4x+4)-(12x^2+8x-15)(2x-4)}{(x^2-4x+4)^2}\\f'(0)=\frac{(24(0)+8)(0^2-4(0)+4)-(12(0)^2+8(0)-15)(2(0)-4)}{(0^2-4(0)+4)^2}\\f'(0)=\frac{(8)(4)-(-15)(-4)}{4^2}\\f'(0)=\frac{8-15}{4}\\\boxed{\boxed{f'(0)=-1\frac{3}{4}}}[/tex]