Find the Local Maxima and Minima e^(1-20x+5x^2)

Problem

e^(1−20*x+5*x^2)

Solution

1. Identify the function as ƒ(x)=e^g(x) where g(x)=5*x^2−20*x+1

2. Apply the chain rule to find the first derivative ƒ(x)^′

d(e^(1−20*x+5*x^2))/d(x)=e^(1−20*x+5*x^2)⋅d(5*x^2−20*x+1)/d(x)

ƒ(x)^′=(10*x−20)*e^(1−20*x+5*x^2)

3. Find the critical points by setting ƒ(x)^′=0 Since e^g(x) is always positive, we solve 10*x−20=0

10*x=20

x=2

4. Apply the second derivative test to determine the nature of the critical point. Find ƒ(x)^″ using the product rule.

ƒ(x)^″=d(10*x−20)/d(x)*e^(1−20*x+5*x^2)+(10*x−20)d(e^(1−20*x+5*x^2))/d(x)

ƒ(x)^″=10*e^(1−20*x+5*x^2)+(10*x−20)^2*e^(1−20*x+5*x^2)

5. Evaluate ƒ(2)^″ to check for concavity.

ƒ(2)^″=10*e^(1−20*(2)+5*(2^2))+(10*(2)−20)^2*e^(1−20*(2)+5*(2^2))

ƒ(2)^″=10*e^(−19)+0

ƒ(2)^″>0

6. Conclude that since ƒ(2)^′=0 and ƒ(2)^″>0 there is a local minimum at x=2

7. Calculate the function value at x=2

ƒ(2)=e^(1−20*(2)+5*(2^2))

ƒ(2)=e^(1−40+20)

ƒ(2)=e^(−19)

Final Answer

Local Minimum: *(2,e^(−19)), Local Maximum: None

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