Find the Local Maxima and Minima f(x)=8/3x^3-2x+1/3

Problem

ƒ(x)=8/3*x3−2*x+1/3

Solution

1. Find the first derivative of the function ƒ(x) using the power rule to determine the slope of the tangent line.

d(ƒ(x))/d(x)=8*x2−2

2. Set the derivative to zero to find the critical points where the slope is horizontal.

8*x2−2=0

3. Solve for x by isolating the variable.

8*x2=2

x2=1/4

x=±1/2

4. Find the second derivative to apply the second derivative test for concavity.

d2(ƒ(x))/(d(x)2)=16*x

5. Evaluate the second derivative at the critical points.

ƒ(1/2)″=16*(1/2)=8

ƒ″*(−1/2)=16*(−1/2)=−8

6. Determine the nature of the points based on the sign of the second derivative. Since ƒ(1/2)″>0 there is a local minimum at x=1/2 Since ƒ″*(−1/2)<0 there is a local maximum at x=−1/2

7. Calculate the y-coordinates by substituting the critical points back into the original function ƒ(x)

ƒ(1/2)=8/3*(1/8)−2*(1/2)+1/3=1/3−1+1/3=−1/3

ƒ*(−1/2)=8/3*(−1/8)−2*(−1/2)+1/3=−1/3+1+1/3=1

Final Answer

Local Maxima: *(−1/2,1), Local Minima: *(1/2,−1/3)

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