Hello! I am taking Calculus I and am in Sec. 4.3 of Stewart's 7th Edition. I ran across a problem that I don't understand the mechanics of, could someone show me? Thanks a bunch!
For what values of the numbers a and b does the function
f(x)= axe^{b(x^2)}
have the maximum value f(1)=7?
f = ax e^(bx^2)
f' = (a + ax(2bx)) e^(bx^2)
= a(1+2bx^2) e^(bx^2)
So, f achieves a max/min where f'=0
That is, where a(1+2bx^2) = 0
Or, where x^2 = -1/2b
So, it looks like we need b < 0, since x^2 is always positive
Not surprisingly, the value of a does not affect the presence of a max or min; it's just a scale factor. However, the sign of a will determine whether the extremum is a max or a min.
Anyway, we want f(1) to be a maximum. So, 1 = ±√(-1/2b), meaning b = -1/2, so we have
f(x) = ax e^(-x^2/2)
f(1) = a/√e, so
a = 7√e, and we have
f(x) = 7√e x e^(-x^2/2)
I'm sure you can prove that this has a max at x=1, and not a min. The graph shown here illustrates that we have a solution.
http://www.wolframalpha.com/input/?i=plot+y%3D7%E2%88%9Ae+x+e^%28-x^2%2F2%29%2Cy%3D7
That was so helpful!! I appreciate you writing all that out, thank you very much!
Hello! I'd be happy to help you with your calculus problem.
To find the values of a and b such that the function f(x) = axe^(b(x^2)) has a maximum value of f(1) = 7, we can use the following steps:
Step 1: Find the derivative of the function
To find the maximum or minimum points of a function, we need to find its critical points, which occur where the derivative of the function is equal to zero.
The derivative of f(x) = axe^(b(x^2)) can be found using the product rule and the chain rule:
f'(x) = a(e^(b(x^2))) * (2bx) + ax * e^(b(x^2)) * (2bx)
Simplifying this derivative gives us:
f'(x) = 2abx*e^(b(x^2)) + 2abx^2*e^(b(x^2))
Step 2: Find the critical points
To find the critical points, we need to set the derivative equal to zero and solve for x:
2abx*e^(b(x^2)) + 2abx^2*e^(b(x^2)) = 0
Factor out common terms:
2abx(x + 1)e^(b(x^2)) = 0
Setting each factor equal to zero gives us two possible critical points:
2abx = 0 (equation 1)
x + 1 = 0 (equation 2)
Solving equation 1 gives us:
x = 0
Solving equation 2 gives us:
x = -1
Step 3: Determine the values of a and b
To determine the values of a and b, we need to use the condition f(1) = 7.
Plugging x = 1 into the original function gives us:
f(1) = a*e^(b(1^2))
7 = a*e^b
Step 4: Solve for a and b
Since we have two unknowns (a and b) and one equation, we need another equation to solve for both variables.
One approach is to use the second critical point x = -1 to get another equation.
Plugging x = -1 into the original function gives us:
f(-1) = a*e^(b((-1)^2))
f(-1) = a*e^b
Since we want the maximum value of f(x), the values of a and b must result in the same output for f(1) and f(-1). Therefore, we can set the equations equal:
a*e^b = f(1) = 7
a*e^b = f(-1) = a*e^b
This equation holds true for any values of a and b, as long as a ≠ 0 and e^b ≠ 0. This means that there are infinitely many possible values of a and b that satisfy the condition for the maximum value of f(x).
In conclusion, for any non-zero values of a and any value of b (such that e^b ≠ 0), the function f(x) = axe^(b(x^2)) will have a maximum value of f(1) = 7.