Find the Inflection Points f(x)=3x(x-3)^3

Problem

ƒ(x)=3*x*(x−3)^3

Solution

1. Find the first derivative using the product rule (d(u)*v)/d(x)=ud(v)/d(x)+vd(u)/d(x)

ƒ(x)^′=3*x⋅d(x−3)/d(x)+(x−3)^3⋅(d(3)*x)/d(x)

ƒ(x)^′=3*x⋅3*(x−3)^2+(x−3)^3⋅3

ƒ(x)^′=9*x*(x−3)^2+3*(x−3)^3

2. Simplify the first derivative by factoring out 3*(x−3)^2

ƒ(x)^′=3*(x−3)^2*(3*x+(x−3))

ƒ(x)^′=3*(x−3)^2*(4*x−3)

3. Find the second derivative using the product rule again.

ƒ(x)^″=3*(x−3)^2⋅d(4*x−3)/d(x)+(4*x−3)⋅(d(3)*(x−3)^2)/d(x)

ƒ(x)^″=3*(x−3)^2⋅4+(4*x−3)⋅6*(x−3)

ƒ(x)^″=12*(x−3)^2+6*(x−3)*(4*x−3)

4. Simplify the second derivative by factoring out 6*(x−3)

ƒ(x)^″=6*(x−3)*(2*(x−3)+(4*x−3))

ƒ(x)^″=6*(x−3)*(2*x−6+4*x−3)

ƒ(x)^″=6*(x−3)*(6*x−9)

ƒ(x)^″=18*(x−3)*(2*x−3)

5. Identify potential inflection points by setting ƒ(x)^″=0

18*(x−3)*(2*x−3)=0

x=3

x=3/2

6. Verify concavity changes by checking the sign of ƒ(x)^″ around the critical values.
   For x<3/2 ƒ(x)^″>0 (concave up).
   For 3/2<x<3 ƒ(x)^″<0 (concave down).
   For x>3 ƒ(x)^″>0 (concave up).
   Since the sign changes at both values, both are inflection points.

7. Calculate the y-coordinates by substituting x values back into ƒ(x)

ƒ(3)=3*(3)*(3−3)^3=0

ƒ(3/2)=3*(3/2)*(3/2−3)^3=9/2*(−3/2)^3=9/2*(−27/8)=−243/16

Final Answer

Inflection Points: *(3/2,−243/16),(3,0)

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