Find the local max/min points for the function y=x^2 * e^x
by the product rule
y' = x^2(e^x) + 2x(e^x)
= e^x(x^2 + 2x) = 0 for max/min
e^x = 0 ---> no solution or
x^2 + 2x = 0
x(x+2) = 0
x = 0 or x = -2
if x=0 then y 0
if x = -2 then y = 4(e^2)
so (0,0) and ( -2,4e^2) are max/min points.
I will let you decide which is which.
To find the local maxima and minima of the function y = x^2 * e^x, we need to take the derivative of the function and find where it equals zero. The points where the derivative equals zero indicate possible extrema.
Let's start by taking the derivative of the function. Remember that we can use the product rule for differentiation when we have a product of functions like this.
Using the product rule, the derivative of y with respect to x would be:
dy/dx = (d/dx)(x^2) * e^x + x^2 * (d/dx)(e^x)
Taking the derivative of each term separately:
(d/dx)(x^2) = 2x (since the derivative of x^n is n * x^(n-1))
(d/dx)(e^x) = e^x (since the derivative of e^x is e^x)
Substituting these derivatives back into our original equation:
dy/dx = 2x * e^x + x^2 * e^x
Now we set the derivative equal to zero and solve for x to find the critical points:
2x * e^x + x^2 * e^x = 0
Factoring out e^x:
e^x * (2x + x^2) = 0
Since e^x is always positive and never zero, for this equation to be true, we must have:
2x + x^2 = 0
This is a quadratic equation, which we can solve by factoring, completing the square, or using the quadratic formula. In this case, we can factor it as:
x(2 + x) = 0
So, we have two solutions:
x = 0 and x = -2
To determine whether these points are local maxima or minima, we need to examine the behavior of the function around these points. We can do this by evaluating the second derivative.
Taking the derivative of the derivative (dy^2/dx^2):
d^2y/dx^2 = (d/dx)(2x * e^x + x^2 * e^x)
Differentiating each term:
(d/dx)(2x * e^x) = 2e^x + 2xe^x
(d/dx)(x^2 * e^x) = 2xe^x + x^2e^x
Adding these derivatives:
d^2y/dx^2 = 2e^x + 2xe^x + 2xe^x + x^2e^x = 4xe^x + x^2e^x
Now we can substitute our critical points x = 0 and x = -2 into this second derivative equation to determine the concavity at these points.
For x = 0:
d^2y/dx^2 = 4(0)*e^0 + (0)^2*e^0 = 0
For x = -2:
d^2y/dx^2 = 4(-2)*e^(-2) + (-2)^2*e^(-2)
= -8e^(-2) + 4e^(-2)
= -4e^(-2)
Since the second derivative is positive when x = 0 and negative when x = -2, we have:
- At x = 0, the concavity changes, so this is a point of inflection, not an extremum.
- At x = -2, the concavity is negative, so it is a local maximum point.
Therefore, the function y = x^2 * e^x has a local maximum point at x = -2.
To summarize:
The local maximum point for the function y = x^2 * e^x is (x, y) = (-2, 4e^(-2)).