What is the equation of the normal line of f(x)=e2x−x2(x−1)2 at x=3?

f(x)=e2x−x2(x−1)2;a=3y=f(a)−1f′(a)(x−a)f(3)=e6−94f′(x)=12(e2x−x2)−12⋅(2e2x−2x)(x−1)2−e2x−x2(2(x−1))(x−1)4f′(x)=12(e2x−x2)−12⋅(2e2x−2x)(x−1)−2e2x−x2(x−1)3f′(3)=(e6−9)−12⋅(2e6−6)−2e6−98≈0.0755y=e6−94−8(x−3)(e6−9)−12⋅(2e6−6)−2e6−9y=-13.24015x+44.6855

"What is the equation of the normal line..." was answered by our community

Destinee Bryan

Answered question

2022-04-10

What is the equation of the normal line of

f (x) = \frac{\sqrt{e^{2 x} - x^{2}}}{{(x - 1)}^{2}}

at x=3?

Answer & Explanation

Marin Lowe

Beginner2022-04-11Added 18 answers

f (x) = \frac{\sqrt{e^{2 x} - x^{2}}}{{(x - 1)}^{2}}; a = 3

y = f (a) - \frac{1}{f^{'} (a)} (x - a)

f (3) = \frac{\sqrt{e^{6} - 9}}{4}

f^{'} (x) = \frac{\frac{1}{2} {(e^{2 x} - x^{2})}^{- \frac{1}{2}} \cdot (2 e^{2 x} - 2 x) {(x - 1)}^{2} - \sqrt{e^{2 x} - x^{2}} (2 (x - 1))}{{(x - 1)}^{4}}

f^{'} (x) = \frac{\frac{1}{2} {(e^{2 x} - x^{2})}^{- \frac{1}{2}} \cdot (2 e^{2 x} - 2 x) (x - 1) - 2 \sqrt{e^{2 x} - x^{2}}}{{(x - 1)}^{3}}

f^{'} (3) = \frac{{(e^{6} - 9)}^{- \frac{1}{2}} \cdot (2 e^{6} - 6) - 2 \sqrt{e^{6} - 9}}{8} \approx 0.0755

y = \frac{\sqrt{e^{6} - 9}}{4} - \frac{8 (x - 3)}{{(e^{6} - 9)}^{- \frac{1}{2}} \cdot (2 e^{6} - 6) - 2 \sqrt{e^{6} - 9}}

y=-13.24015x+44.6855

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