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

Problem

ƒ(x)=e^(1−5*x^2)

Solution

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

d(e^(1−5*x^2))/d(x)=e^(1−5*x^2)⋅d(1−5*x^2)/d(x)

ƒ(x)^′=e^(1−5*x^2)⋅(−10*x)

ƒ(x)^′=−10*x*e^(1−5*x^2)

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

−10*x*e^(1−5*x^2)=0

x=0

Note: e^(1−5*x^2) 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*x^2)+(−10*x)⋅d(e^(1−5*x^2))/d(x)

ƒ(x)^″=−10*e^(1−5*x^2)+(−10*x)*(−10*x*e^(1−5*x^2))

ƒ(x)^″=−10*e^(1−5*x^2)+100*x^2*e^(1−5*x^2)

ƒ(x)^″=10*e^(1−5*x^2)*(10*x^2−1)

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*e^1*(−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)=e^1=e

Final Answer

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

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