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

Problem

ƒ(x)=3*x4+4*x3

Solution

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

d(ƒ(x))/d(x)=12*x3+12*x2

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

d2(ƒ(x))/(d(x)2)=36*x2+24*x

3. Set the second derivative to zero to find the potential inflection points.

36*x2+24*x=0

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

12*x*(3*x+2)=0

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

x=0

x=−2/3

6. Test for concavity changes by checking the sign of ƒ(x)″ in the intervals (−∞,−2/3) (−2/3,0) and (0,∞)

ƒ″*(−1)=36*(−1)2+24*(−1)=12>0

ƒ″*(−0.5)=36*(−0.5)2+24*(−0.5)=9−12=−3<0

ƒ(1)″=36*(1)2+24*(1)=60>0

7. Calculate the y-coordinates for the points where the concavity changes.

ƒ(0)=3*(0)4+4*(0)3=0

ƒ*(−2/3)=3*(−2/3)4+4*(−2/3)3=3*(16/81)+4*(−8/27)=16/27−32/27=−16/27

Final Answer

Inflection Points:(0,0),(−2/3,−16/27)

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