What is the equation of the line that is normal to&nbsp;f(x)=-3x2-sinx&nbsp;at&nbsp;x=π3?

f(x)=−3x2−sin⁡xa=π3n:y=f(a)−1f′(a)⋅(x−a)f(π3)=−(3(π3)2)−sin⁡(π3)=−π23−32f′(x)=−6x−cos⁡xf′(π3)=−6π3−cos⁡(π3)=−2π−12=−4π−12=−4π+12n:y=−π23−32−1−4π+12⋅(x−π3)n:y=−π23−32+24π+1⋅(x−π3)

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Jakobe Norton

Answered question

2022-04-09

What is the equation of the line that is normal to $f (x) = - 3 x^{2} - \sin x$ at $x = \frac{π}{3}$ ?

Answer & Explanation

tempur8x43

Beginner2022-04-10Added 16 answers

f (x) = - 3 x^{2} - \sin x

a = \frac{π}{3}

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

f (\frac{π}{3}) = - (3 {(\frac{π}{3})}^{2}) - \sin (\frac{π}{3}) = - \frac{π^{2}}{3} - \frac{\sqrt{3}}{2}

f^{'} (x) = - 6 x - \cos x

f^{'} (\frac{π}{3}) = - 6 \frac{π}{3} - \cos (\frac{π}{3}) = - 2 π - \frac{1}{2} = \frac{- 4 π - 1}{2} = - \frac{4 π + 1}{2}

n : y = - \frac{π^{2}}{3} - \frac{\sqrt{3}}{2} - \frac{1}{- \frac{4 π + 1}{2}} \cdot (x - \frac{π}{3})

n : y = - \frac{π^{2}}{3} - \frac{\sqrt{3}}{2} + \frac{2}{4 π + 1} \cdot (x - \frac{π}{3})

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