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

Problem

ƒ(x)=e(1−2*x2)

Solution

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

d(e(1−2*x2))/d(x)=e(1−2*x2)⋅d(1−2*x2)/d(x)

ƒ(x)′=e(1−2*x2)⋅(−4*x)

ƒ(x)′=−4*x*e(1−2*x2)

2. Identify critical points by setting the first derivative equal to zero.

−4*x*e(1−2*x2)=0

x=0

3. Find the second derivative using the product rule to test the nature of the critical point.

ƒ(x)″=d(−4*x)/d(x)⋅e(1−2*x2)+(−4*x)⋅d(e(1−2*x2))/d(x)

ƒ(x)″=−4*e(1−2*x2)+(−4*x)*(−4*x*e(1−2*x2))

ƒ(x)″=−4*e(1−2*x2)+16*x2*e(1−2*x2)

ƒ(x)″=e(1−2*x2)*(16*x2−4)

4. Apply the second derivative test by substituting the critical point x=0 into ƒ(x)″

ƒ(0)″=e(1−2*(0)2)*(16*(0)2−4)

ƒ(0)″=e1*(−4)

ƒ(0)″=−4*e

Since ƒ(0)″<0 a local maximum occurs at x=0

5. Calculate the y-coordinate of the local maximum.

ƒ(0)=e(1−2*(0)2)

ƒ(0)=e1

ƒ(0)=e

Final Answer

Local Maximum: *(0,e), Local Minimum: None

Want more problems? Check here!