Find a formula for a curve of the form y=e−(x−a)2/b
for b>0 with a local maximum at x=6 and points of inflection at x=3 and x=9.
If you want y = e^(-(x-a)^2/b) then
y' = -1/b (2(x-a)) e^(-(x-a)^2/b)
y" = [-1/b (2) + -1/b (2(x-a)) [-1/b (2(x-a))]] e^(-(x-a)^2/b)
= (4(x-a)^2/b^2 - 2/b) e^(-(x-a)^2/b)
= 2/b^2 (2(x-a)^2 - b) e^(-(x-a)^2/b)
So, y"=0 at x=a±√(b/2)
3 = 6-3
9 = 6+3
so we need a=6, b=12
That gives us
y = e^(-(x-6)^2/12)
see the plot at
wolframalpha. com/input?i=+e%5E%28-%28x-6%29%5E2%2F12%29
To find a formula for the curve of the given form, we need to determine the values of the parameters "a" and "b" that satisfy the given conditions.
Let's start by analyzing the given information:
1. A local maximum at x = 6:
For a local maximum to occur at x = 6, the derivative of the function must be zero at that point. So, we need to find the derivative of the function and set it to zero.
Given function: y = e^(-(x - a)^2 / b)
Taking the derivative of "y" with respect to "x":
dy/dx = (-2(x - a) / b) * e^(-(x - a)^2 / b)
Setting the derivative equal to zero and solving for "x":
0 = (-2(x - a) / b) * e^(-(x - a)^2 / b)
Since e^(-(x - a)^2 / b) is always positive, we can ignore it, and the equation simplifies to:
0 = -2(x - a)
Solving for "x", we get:
x = a
Therefore, the parameter "a" is equal to 6.
2. Points of inflection at x = 3 and x = 9:
For points of inflection to occur at x = 3 and x = 9, the second derivative of the function must equal zero at those points. So, we need to find the second derivative of the function and set it to zero.
Taking the derivative of the first derivative:
d²y/dx² = (2(b - (x - a)^2)) * e^(-(x - a)^2 / b) / b^2
Setting the second derivative equal to zero and solving for "x":
0 = (2(b - (x - a)^2)) * e^(-(x - a)^2 / b) / b^2
Again, since e^(-(x - a)^2 / b) is always positive, we can ignore it, and the equation simplifies to:
0 = 2(b - (x - a)^2)
Expanding and simplifying:
0 = 2b - 2(x - a)^2
0 = b - (x - a)^2
To find "b", we substitute the values of x from the points of inflection:
0 = b - (3 - 6)^2
0 = b - 9
Solving for "b", we get:
b = 9
So, the formula for the curve of the given form with a local maximum at x = 6 and points of inflection at x = 3 and x = 9 is:
y = e^(-(x - 6)^2 / 9)
To find the formula for the given curve, we will use the information provided:
1. Local maximum at x = 6:
To have a local maximum at x = 6, the derivative of the function must be zero at x = 6, and the second derivative must be negative. Let's calculate:
First derivative: y' = d/dx(e^(-(x-a)^2/b)) = -2(x-a)e^(-(x-a)^2/b)/b
Setting y' = 0 and solving for x:
0 = -2(x-a)e^(-(x-a)^2/b)/b
Since b > 0, the exponential term is never zero, so we have:
0 = -2(x-a) => x = a
Therefore, a = 6.
2. Points of inflection at x = 3 and x = 9:
To have points of inflection at x = 3 and x = 9, the second derivative must be zero at those points. Let's calculate:
Second derivative: y'' = d^2/dx^2(e^(-(x-a)^2/b)) = (2(x-a)^2/b - 1)e^(-(x-a)^2/b)/b^2
Setting y'' = 0 and solving for x:
0 = (2(x-a)^2/b - 1)e^(-(x-a)^2/b)/b^2
Again, since b > 0, the exponential term is never zero. We can ignore the denominator b^2. Solving for x, we have:
2(x-a)^2 = b
For x = 3, plugging into the equation above gives:
2(3-6)^2 = b => 2(9) = b => b = 18
For x = 9, plugging into the equation above gives:
2(9-6)^2 = b => 2(9) = b => b = 18
Therefore, the value of b is 18.
Now that we know a = 6 and b = 18, we can substitute these values into the original equation to obtain the final formula for the curve:
y = e^(-(x-6)^2/18)