Find the Inflection Points f(x)=1/2x^4+2x^3

Problem

ƒ(x)=1/2*x4+2*x3

Solution

1. Find the first derivative by applying the power rule to each term of the function.

d(ƒ(x))/d(x)=2*x3+6*x2

2. Find the second derivative by differentiating the first derivative with respect to x

d2(ƒ(x))/(d(x)2)=6*x2+12*x

3. Set the second derivative to zero to find potential inflection points where the concavity might change.

6*x2+12*x=0

4. Factor the equation to solve for the critical values of x

6*x*(x+2)=0

5. Solve for x by setting each factor to zero.

x=0

x=−2

6. Test the intervals around the critical values in the second derivative to confirm a change in concavity.
   For x<−2 ƒ″*(−3)=18>0 (concave up).
   For −2<x<0 ƒ″*(−1)=−6<0 (concave down).
   For x>0 ƒ(1)″=18>0 (concave up).
   Since the sign changes at both x=−2 and x=0 both are inflection points.

7. Find the y-coordinates by substituting the x values back into the original function ƒ(x)

ƒ*(−2)=1/2*(−2)4+2*(−2)3=8−16=−8

ƒ(0)=1/2*(0)4+2*(0)3=0

Final Answer

Inflection Points: *(−2,−8),(0,0)

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