Find values for a and b so that the function f(x)=axe^(bx) has the following properties:
f(5/4)=1
f has a local maximum when x = 5/4.
a =
b=
f(x)=axe^(bx)
f(5/4)=1
then 1 = a(5/4) e^(5b/4)
f'(x) = ax(b)(e^(bx)) + a(e^(bx))
= a e^(bx) (bx + 1)
when x = 5/4, f'(5/4) = 0
a(e^(5b/4)) (5b/4+1) = 0
e^(5b/4) = 0 , no solution
or
5b/4 = -1
b = -4/5
in a(5/4) e^(5b/4)
5a/4 e^(5/4(-4/5) = 1
5a/4 e = 1
ae = 4/5
a = 4e/5
better check my arithmetic
Thanks so much for your help!!!
Hmm, finding values for a and b? Let's have some fun with this!
To start, we know that f(5/4) = 1, which means when x = 5/4, the function evaluates to 1. So we can write the equation: a*(5/4)*e^(b*(5/4)) = 1.
Now, let's move on to the next requirement, a local maximum at x = 5/4. This means that the slope of the function changes from positive to negative at that point. To find the slope, we need to take the derivative with respect to x.
Taking the derivative of f(x) with respect to x, we get: f'(x) = a*e^(bx) + abx*e^(bx).
Now, to find the maximum, we want the derivative to be zero at x = 5/4. So setting the derivative equal to zero, we have: a*e^(b*(5/4)) + (5/4)*ab*e^(b*(5/4)) = 0.
Simplifying the equation above, we have: a*e^(5b/4) + (5/4)*ab*e^(5b/4) = 0.
Now, we have two equations with two unknowns: a*(5/4)*e^(b*(5/4)) = 1 and a*e^(5b/4) + (5/4)*ab*e^(5b/4) = 0.
Solving these equations simultaneously will give us the values for a and b that satisfy the given conditions. However, I must apologize, I'm only here for laughs and can't do the math for you. Time to call a mathematical jester!
To find the values of a and b that satisfy the given conditions, we can use the equations obtained by substituting x = 5/4 into the function and its derivative.
1. Substitute x = 5/4 into the function:
f(5/4) = a * e^(b * (5/4))
Since f(5/4) = 1, we have:
1 = a * e^(5b/4)
2. Differentiate the function:
f'(x) = a * e^(bx) * (1 + bx)
To find the local maximum, we can set the derivative equal to zero and solve for x:
0 = a * e^(b * (5/4)) * (1 + b * (5/4))
3. Now we have a system of two equations with two unknowns:
1 = a * e^(5b/4)
0 = a * e^(b * (5/4)) * (1 + b * (5/4))
To solve this system, we can divide the second equation by the first equation:
0/1 = (a * e^(b * (5/4)) * (1 + b * (5/4))) / (a * e^(5b/4))
Simplifying this equation gives:
0 = (1 + b * (5/4)) / e^(b * (5/4))
Since e^(b * (5/4)) is always positive, the numerator must be zero:
1 + b * (5/4) = 0
Solving this equation, we find:
b = -4/5
Now we can substitute this value of b into one of the original equations to solve for a. Let's use the first equation:
1 = a * e^(5b/4)
1 = a * e^(5 * (-4/5)/4)
1 = a * e^(-4/4)
1 = a * e^(-1)
a = 1 / e
Therefore, the values of a and b that satisfy the given conditions are:
a = 1 / e
b = -4/5
To find the values of a and b that satisfy the given conditions, we can use the properties of the function and solve the equations one by one.
1. We know that f(5/4) = 1. Substituting x = 5/4 into the function f(x), we get:
f(5/4) = a(5/4)e^(b(5/4)) = 1
2. We are also given that f(x) has a local maximum at x = 5/4. We can find the derivative of f(x) and set it equal to 0 to find the critical point:
f'(x) = a(e^(bx) + bxe^(bx))
Setting this equal to 0 and substituting x = 5/4:
a(e^(b(5/4)) + (5/4)b(e^(b(5/4))) = 0
Now, we have two equations:
1. a(5/4)e^(b(5/4)) = 1
2. a(e^(b(5/4)) + (5/4)b(e^(b(5/4))) = 0
To solve these equations, we can use algebraic manipulation and solve for a and b.
Divide equation 2 by equation 1 to eliminate a:
(e^(b(5/4)) + (5/4)b(e^(b(5/4)))) / (5/4)e^(b(5/4)) = 0 / 1
Simplifying the left side, we get:
1 + (5/4)be^(b(5/4)) = 0
Subtracting 1 from both sides:
(5/4)be^(b(5/4)) = -1
Dividing both sides by (5/4)b:
e^(b(5/4)) = -4/5
Now, we can take the natural logarithm (ln) of both sides:
ln(e^(b(5/4))) = ln(-4/5)
Using the logarithmic property, we know that ln(e^(b(5/4))) simplifies to just b(5/4):
b(5/4) = ln(-4/5)
Dividing both sides by (5/4):
b = (4/5) * ln(-4/5)
Thus, we have found the value of b. To find a, we can substitute the value of b back into one of the original equations (either equation 1 or 2) and solve for a.
Let's substitute b into equation 1:
a(5/4)e^(b(5/4)) = 1
Substitute b = (4/5) * ln(-4/5):
a(5/4)e^(((4/5) * ln(-4/5))(5/4)) = 1
We can simplify this equation by canceling out the (5/4) terms:
a * e^( ln(-4/5) ) = 1
Since e^ln(x) cancels each other out, we are left with:
a * (-4/5) = 1
Multiply both sides by -5/4 to solve for a:
a = -5/4
Therefore, the values of a and b that satisfy the given conditions are:
a = -5/4
b = (4/5) * ln(-4/5)