What is the equation of the line normal to f(x)=x3ex at x=2?

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Wernbergbo9d · Accepted Answer

Explanation:The equation of the line normal to the graph of the function:y=f(x)at ath point (x―,f(x―)) is given by:y=f(x―)−1f′(x―)(x−x―)In our case:f(x)=x3ex=x3e−xf(2)=8e2f′(x)=3x2e−x−x3e−xf′(2)=12e2−8e2=4e2So the normal line is:y=8e2−e24(x−2)

tralhavahr9c · Accepted Answer

We are given that f(x)=x3e−x and x0=2Find the value of the function at the given point: y0=f(2)=8e2The slope of the normal line at x=x0 is the negative reciprocal of the derivative of the function,evaluated at x=x0:M(x0)=−1f′(x0)Find the derivative: f′(x)=(x3e−x)′=x2(3−x)e−xHence, M(x0)=−1f′(x0)=−ex0x02(3−x0)Next, find the slope at the given point.m=M(2)=−e24Finally, the equation of the normal line is y−y0=m(x−x0)Plugging the found values, we get that y−8e2=−e24(x−2)Or, more simply: y=−xe24+16+e42e2

What is the equation of the line normal

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