Find constants a and b in the function f(x)=axe^(bx)such that f(3/4)=1 and the function has a local maximum at x=3/4.
a=
b=
y = a x e^(bx)
dy/dx = ax [ b e^(bx ] + a e^(bx) = a e^(bx) [ x b + 1]
that is zero at x = 3/4
so (3/4)b + 1 = 0
b = -4/3
y =a x e^-(4x/3)
1 = a (3/4) e^- 1
e = (3/4) a
a = 4 e / 3
a=3.31612560229033
b=-4.46818076947058
To find the constants a and b in the function f(x) = axe^(bx), we are given that f(3/4) = 1 and that the function has a local maximum at x = 3/4.
Step 1: Find the function's derivative.
Taking the derivative of f(x) with respect to x, we get:
f'(x) = a * e^(bx) + abx * e^(bx)
Step 2: Set up the equations using the given information.
We need to find a and b such that f(3/4) = 1. This means plugging x = 3/4 into the function should equal 1.
f(3/4) = a * e^(b(3/4)) = 1 Equation 1
We also know that the function has a local maximum at x = 3/4. At this point (x = 3/4), the derivative f'(x) = 0.
f'(3/4) = a * e^(b(3/4)) + ab(3/4) * e^(b(3/4)) = 0 Equation 2
Step 3: Solve the system of equations.
We now have two equations (Equation 1 and Equation 2) with two unknowns (a and b). We can solve this system of equations to find the values of a and b.
From Equation 2, we can factor out e^(b(3/4)):
e^(b(3/4)) * (a + (3/4)ab) = 0
Since e^(b(3/4)) is always positive and non-zero, we can set the expression in the parentheses equal to 0:
a + (3/4)ab = 0 Equation 3
To solve for a, we can substitute Equation 3 into Equation 1:
(3/4)a + (9/16)a^2 = 1
Multiplying both sides by 16:
12a + 9a^2 = 16
Rearranging the equation:
9a^2 + 12a - 16 = 0
Step 4: Solve the quadratic equation.
We now have a quadratic equation in terms of a. Solving this equation can give us the possible values of a.
Using the quadratic formula:
a = (-b ± √(b^2 - 4ac)) / (2a)
In our case, the coefficients are:
a = 9
b = 12
c = -16
Applying these values to the quadratic formula, we get:
a = (-12 ± √((12)^2 - 4(9)(-16))) / (2(9))
Simplifying:
a = (-12 ± √(144 + 576)) / 18
a = (-12 ± √720) / 18
Further simplifying:
a = (-12 ± 12√5) / 18
This can be further simplified as:
a = (-2 ± 2√5) / 3
So the possible values of a are (-2 + 2√5) / 3 and (-2 - 2√5) / 3.
Step 5: Find b.
To find the value of b, we can substitute the values of a into Equation 3:
a + (3/4)ab = 0
Substituting the values, we get:
[(-2 + 2√5) / 3] + (3/4) * [(-2 + 2√5) / 3] * b = 0
We can simplify this equation and solve for b.
Simplifying:
[(-2 + 2√5) / 3] + [(-1 + √5)/2] * (b) = 0
Multiplying both sides by 3:
-2 + 2√5 + [-1 + √5] * (b) = 0
Rearranging the equation:
[-1 + √5] * (b) = 2 - 2√5
Dividing both sides by [-1 + √5]:
b = (2 - 2√5) / [-1 + √5]
So the value of b is (2 - 2√5) / (-1 + √5).
Therefore, the values of a and b in the function f(x) = axe^(bx) such that f(3/4) = 1 and the function has a local maximum at x = 3/4 are:
a = (-2 + 2√5) / 3
b = (2 - 2√5) / (-1 + √5)
To find the values of constants a and b in the function f(x) = axe^(bx) such that f(3/4) = 1 and the function has a local maximum at x = 3/4, we can use calculus and algebraic manipulation.
Step 1: Evaluate f(3/4) = 1
Substitute x = 3/4 in the function f(x) = axe^(bx):
f(3/4) = a(3/4)e^(b(3/4)) = 1
Step 2: Find the derivative of f(x)
Calculate the first derivative of f(x) with respect to x:
f'(x) = (a + abx)e^(bx)
Step 3: Set up an equation for the local maximum
To have a local maximum at x = 3/4, we need f'(3/4) = 0. So, set up the equation:
f'(3/4) = (a + ab(3/4))e^(b(3/4)) = 0
Step 4: Solve the equations
We now have two equations:
a(3/4)e^(b(3/4)) = 1 (from Step 1)
(a + ab(3/4))e^(b(3/4)) = 0 (from Step 3)
Since the second equation is 0, we can ignore it and focus on the first equation:
a(3/4)e^(b(3/4)) = 1
To solve for a and b, we can proceed as follows:
Divide both sides of the equation by (3/4)e^(b(3/4)):
a = (4/3)e^(-b(3/4))
Now, substitute this value of a into the first equation:
(4/3)e^(-b(3/4))(3/4)e^(b(3/4)) = 1
Simplify:
(4/3)(3/4)e^(b(3/4) - b(3/4)) = 1
e^0 = 1
1 = 1
Since 1 = 1 is a true statement, it means that any value of a and b can satisfy the conditions f(3/4) = 1 and having a local maximum at x = 3/4. This implies that there are infinitely many possible values for the constants a and b that fulfill these conditions.