Find the Local Maxima and Minima y=x^4-2x^2

Problem

y=x4−2*x2

Solution

1. Find the first derivative of the function with respect to x to locate critical points.

d(y)/d(x)=4*x3−4*x

2. Set the derivative to zero and solve for x to find the critical values.

4*x3−4*x=0

4*x*(x2−1)=0

4*x*(x−1)*(x+1)=0

x=0,x=1,x=−1

3. Find the second derivative to apply the Second Derivative Test for concavity.

d2(y)/(d(x)2)=12*x2−4

4. Evaluate the second derivative at each critical point to determine if it is a maximum or minimum.

At *x=0:d2(y)/(d(x)2)=12*(0)2−4=−4

At *x=1:d2(y)/(d(x)2)=12*(1)2−4=8

At *x=−1:d2(y)/(d(x)2)=12*(−1)2−4=8

5. Interpret the results where a negative second derivative indicates a local maximum and a positive second derivative indicates a local minimum.

Local maximum at *x=0

Local minima at *x=1* and *x=−1

6. Calculate the y-coordinates by substituting the x values back into the original function.

y(0)=(0)4−2*(0)2=0

y(1)=(1)4−2*(1)2=−1

y*(−1)=(−1)4−2*(−1)2=−1

Final Answer

Local Maxima: *(0,0), Local Minima: *(1,−1),(−1,−1)

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