the function f is defined as f(x)=\\frac{7}{2x^{2}-4x}.\nfind f(x + 6).\nwrite your answer without…

Explanation:

Step1: Find the derivative of (f(x))

Use the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}). Here (u = 7), (u^\prime=0), (v = 2x^{2}-4x), (v^\prime = 4x - 4). [ \begin{align*} f^\prime(x)&=\frac{0\times(2x^{2}-4x)-7\times(4x - 4)}{(2x^{2}-4x)^{2}}\ &=\frac{-28x + 28}{(2x^{2}-4x)^{2}} \end{align*} ]

Step2: Substitute (x) with (x + 6)

Replace (x) in (f^\prime(x)) with (x+6). For the numerator: (-28(x + 6)+28=-28x-168 + 28=-28x-140) For the denominator: ([2(x + 6)^{2}-4(x + 6)]^{2}) First, expand (2(x + 6)^{2}-4(x + 6)): [ \begin{align*} 2(x^{2}+12x + 36)-4x-24&=2x^{2}+24x+72-4x - 24\ &=2x^{2}+20x + 48 \end{align*} ] So (f^\prime(x + 6)=\frac{-28x-140}{(2x^{2}+20x + 48)^{2}}) Factor numerator and denominator: Numerator: (-28x-140=-28(x + 5)) Denominator: (2x^{2}+20x + 48=2(x^{2}+10x + 24)=2(x + 4)(x+6)), ((2x^{2}+20x + 48)^{2}=4(x + 4)^{2}(x + 6)^{2}) [ f^\prime(x + 6)=\frac{-28(x + 5)}{4(x + 4)^{2}(x + 6)^{2}}=\frac{-7(x + 5)}{(x + 4)^{2}(x + 6)^{2}} ]

Answer:

(\frac{-7(x + 5)}{(x + 4)^{2}(x + 6)^{2}})