Find the Local Maxima and Minima f(x)=e^(1-5x^2)

Problem

ƒ(x)=e(1−5*x2)

Solution

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

d(e(1−5*x2))/d(x)=e(1−5*x2)⋅d(1−5*x2)/d(x)

ƒ(x)′=e(1−5*x2)⋅(−10*x)

ƒ(x)′=−10*x*e(1−5*x2)

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

−10*x*e(1−5*x2)=0

x=0

Note: e(1−5*x2) is always positive and never zero.

3. Find the second derivative using the product rule to apply the Second Derivative Test.

ƒ(x)″=d(−10*x)/d(x)⋅e(1−5*x2)+(−10*x)⋅d(e(1−5*x2))/d(x)

ƒ(x)″=−10*e(1−5*x2)+(−10*x)*(−10*x*e(1−5*x2))

ƒ(x)″=−10*e(1−5*x2)+100*x2*e(1−5*x2)

ƒ(x)″=10*e(1−5*x2)*(10*x2−1)

4. Evaluate the second derivative at the critical point x=0

ƒ(0)″=10*e(1−5*(0)2)*(10*(0)2−1)

ƒ(0)″=10*e1*(−1)

ƒ(0)″=−10*e

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

5. Calculate the function value at the local maximum.

ƒ(0)=e(1−5*(0)2)

ƒ(0)=e1=e

Final Answer

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

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