Consider the function f(x)=x^n e^(-2x) for x >/= 0, n > 2
A. Find the constant n for which the function f(x) attains its maximum value at x=3.
B. For n=4, find all x values of points of inflection for the curve y=f(x)
f(x)= (x^n)( e^(-2x))
f'(x) = x^n (-2)(e^(-2x) + e^(-2x) (n)x^(n-1)
= 0 when x = 3
e^(-2x) ( -2x^n + nx^(n-1) ) = 0
but x = 3, so
e^-6 (-2(3)^n + n(3)^(n-1) ) = 0
divide out the e^-6
(-2(3)^n + n(3)^(n-1) ) = 0
3^(n-1) (-2(3) + n) = 0
3^(n-1) = 0 ----> so solution
or
-6 + n = 0
n = 6
b)
take the derivative of my f'(x), set it equal to zero, and sub in x = 4
then solve for n
go for it!
A. To find the constant n for which the function f(x) attains its maximum value at x=3, we need to find the critical points of f(x) and determine the value of n that satisfies the condition.
Let's start by finding the derivative of f(x) with respect to x:
f'(x) = n*x^(n-1) * e^(-2x) - 2*x^n * e^(-2x)
To find the critical points, we need to set the derivative equal to zero and solve for x:
n*x^(n-1) * e^(-2x) - 2*x^n * e^(-2x) = 0
Since we're looking for x=3, we can substitute that value into the equation and solve for n:
n*3^(n-1) * e^(-6) - 2*3^n * e^(-6) = 0
I'm starting to doubt if this bot can solve the equation. Sorry!
A. To find the constant n for which the function f(x) attains its maximum value at x = 3, we need to find the critical point of the function. The critical point occurs when the first derivative of the function is equal to zero.
Let's find the first derivative of f(x) with respect to x:
f'(x) = d/dx [x^n e^(-2x)]
Utilizing the product rule, we obtain:
f'(x) = n(x^(n-1))e^(-2x) + x^n(-2)e^(-2x)
To find the critical point, set f'(x) equal to zero and solve for x:
0 = n(x^(n-1))e^(-2x) + x^n(-2)e^(-2x)
Simplifying the equation by dividing through by e^(-2x), we get:
0 = n(x^(n-1)) - 2x^n
Next, factor out the common term x^(n-1):
0 = x^(n-1)(n - 2x)
Since x cannot be zero (as per the given condition x >= 0), we can divide both sides of the equation by x^(n-1) to obtain:
0 = n - 2x
Solving for x, we find:
x = n/2
Since we want x = 3 to maximize the function, we have:
3 = n/2
Simplifying, we find:
n = 6
Therefore, the constant n for which the function f(x) attains its maximum value at x = 3 is n = 6.
B. Given n = 4, we need to find all x-values of points of inflection for the curve y = f(x).
For a point of inflection to occur, the second derivative of the function must change sign. So, we need to find where the second derivative changes sign.
Let's find the second derivative of f(x) with respect to x:
f''(x) = d^2/dx^2 [x^4 e^(-2x)]
Utilizing the product rule again, we obtain:
f''(x) = 4(x^3)e^(-2x) + 4(x^4)(-2)e^(-2x) + (x^4)(-2)e^(-2x)
Simplifying further, we get:
f''(x) = 4x^3e^(-2x) - 8x^4e^(-2x) - 2x^4e^(-2x)
Now, set f''(x) equal to zero and solve for x:
0 = 4x^3e^(-2x) - 8x^4e^(-2x) - 2x^4e^(-2x)
Factoring out the common term x^3e^(-2x), we have:
0 = x^3e^(-2x)(4 - 8x - 2x)
Simplifying further, we get:
0 = x^3e^(-2x)(4 - 10x)
This equation will be satisfied if any of the factors on the left side are equal to zero. Therefore, we have two separate cases to consider:
Case 1: x^3e^(-2x) = 0
This occurs when x = 0 (from the given condition x ≥ 0).
Case 2: 4 - 10x = 0
Solving for x, we find:
x = 4/10
Simplifying, we get:
x = 2/5
Therefore, for n = 4, the x-values of points of inflection for the curve y = f(x) are x = 0 and x = 2/5.
To find the constant n for which the function f(x) attains its maximum value at x=3, we need to maximize the function by finding its critical points.
A critical point is a point where the derivative of the function is either zero or undefined. To find the critical points of f(x), we need to find its first derivative, set it equal to zero, and solve for x.
Let's start by finding the first derivative of f(x):
f'(x) = (d/dx) [x^n e^(-2x)]
To do this, we can use the product rule for differentiation. The product rule states that for two functions u(x) and v(x), the derivative of their product is given by:
(d/dx) [u(x)v(x)] = u'(x)v(x) + v'(x)u(x)
For f(x) = x^n e^(-2x), we can let u(x) = x^n and v(x) = e^(-2x). Applying the product rule:
f'(x) = u'(x)v(x) + v'(x)u(x)
= n x^(n-1) e^(-2x) + e^(-2x) x^n (-2)
Simplifying further:
f'(x) = e^(-2x) x^(n-1) [n - 2x]
To find the critical points, we set the derivative f'(x) equal to zero and solve for x:
e^(-2x) x^(n-1) [n - 2x] = 0
Since e^(-2x) is always positive for x ≥ 0, we can neglect it and focus on the expression x^(n-1) [n - 2x]. Setting this expression equal to zero:
x^(n-1) [n - 2x] = 0
Either x^(n-1) = 0 or n - 2x = 0
For x^(n-1) = 0, this means x = 0 (as long as n > 1).
For n - 2x = 0, we find x = n/2 (as long as n ≠ 0).
Therefore, the critical points for f(x) are x = 0 and x = n/2 (for n ≠ 0).
Since we are interested in finding the constant n for which f(x) attains its maximum value at x = 3, we set x = 3 in the expression for f(x):
f(3) = 3^n e^(-2*3) = 3^n e^(-6)
Now we can solve for the constant n that maximizes f(3).