Find constants a and b in the function f(x)=axe^(bX) such that f(1/8)=1 and the function has a local maximum at x=1/8.
f = ax e^(bx)
f' = ae^(bx) + abx e^(bx)
= ae^(bx) (1+bx)
f'(1/8) = 0, so b = -8
f(1/8) = 1 = a(1/8)e^(-1), so a=8e
f(x) = (8e)x e^(-8x)
To find the constants a and b in the function f(x) = axe^(bx) given that f(1/8) = 1 and the function has a local maximum at x = 1/8, we need to use the information given and solve the equations step by step.
Step 1: Start with the function f(x) = axe^(bx) and substitute x = 1/8 and f(1/8) = 1.
1 = a(1/8)e^(b(1/8))
Step 2: Simplify the equation by multiplying through by 8e^(b/8).
8e^(b/8) = a*e^(b/8)
Step 3: Divide both sides by e^(b/8) to isolate a.
8 = a
Now we have the value of a.
Step 4: Use the derivative of f(x) to find the value of b.
Let's differentiate f(x) with respect to x:
f'(x) = a * (e^(bx)) * b + ax * (e^(bx)) * b
To find the local maximum, we need to set the derivative equal to zero:
0 = a * (e^(b(1/8))) * b + a*(1/8) * (e^(b(1/8))) * b
Step 5: Simplify the equation by dividing through by a * (e^(b/8)).
0 = b + (1/8) * b
Step 6: Combine like terms:
0 = (9/8) * b
Step 7: Solve for b:
b = 0
Now we have the value of b.
Therefore, the constants a and b in the function f(x) = axe^(bx) such that f(1/8) = 1 and the function has a local maximum at x = 1/8 are a = 8 and b = 0.
you suck steve
Calculus - Steve, Wednesday, March 6, 2013 at 5:00pm
f = ax e^(bx)
f' = ae^(bx) + abx e^(bx)
= ae^(bx) (1+bx)
f'(1/8) = 0, so b = -8
f(1/8) = 1 = a(1/8)e^(-1), so a=8e
f(x) = (8e)x e^(-8x)