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

Problem

ƒ(x)=(x2−1)3

Solution

1. Find the first derivative of the function using the power rule and the chain rule.

d(ƒ(x))/d(x)=3*(x2−1)2⋅2*x

d(ƒ(x))/d(x)=6*x*(x2−1)2

2. Identify critical points by setting the first derivative equal to zero and solving for x

6*x*(x2−1)2=0

x=0

x2−1=0⇒x=1,x=−1

3. Find the second derivative to apply the Second Derivative Test, using the product rule on 6*x*(x2−1)2

d2(ƒ(x))/(d(x)2)=6*(x2−1)2+6*x⋅2*(x2−1)⋅2*x

d2(ƒ(x))/(d(x)2)=6*(x2−1)2+24*x2*(x2−1)

4. Evaluate the second derivative at the critical points.

At *x=0:6*(0−1)2+24*(0)2*(0−1)=6

At *x=1:6*(1−1)2+24*(1)2*(1−1)=0

At *x=−1:6*(−1−1)2+24*(−1)2*(−1−1)=0

5. Apply the First Derivative Test for points where the second derivative is zero (x=1 and x=−1.

For *x=−1:sign of *ƒ′*(−2)=(−)*(+)=− and sign of *ƒ′*(−0.5)=(−)*(+)=−

For *x=1:sign of *ƒ(0.5)′=(+)*(+)=+ and sign of *ƒ(2)′=(+)*(+)=+

Since the sign of the derivative does not change at x=−1 or x=1 these are points of inflection, not local extrema.

6. Determine the extremum at x=0 Since ƒ(0)″=6>0 the function has a local minimum at x=0

ƒ(0)=(0−1)3=−1

Final Answer

Local Minimum: *(0,−1), Local Maxima: None

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