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

Problem

e(1−20*x+5*x2)

Solution

1. Identify the function as ƒ(x)=eg(x) where g(x)=5*x2−20*x+1

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

d(e(1−20*x+5*x2))/d(x)=e(1−20*x+5*x2)⋅d(5*x2−20*x+1)/d(x)

ƒ(x)′=(10*x−20)*e(1−20*x+5*x2)

3. Find the critical points by setting ƒ(x)′=0 Since eg(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*x2)+(10*x−20)d(e(1−20*x+5*x2))/d(x)

ƒ(x)″=10*e(1−20*x+5*x2)+(10*x−20)2*e(1−20*x+5*x2)

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

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

ƒ(2)=e(1−40+20)

ƒ(2)=e(−19)

Final Answer

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

Want more problems? Check here!