Maximum or minimum values of the function f(x) = (1 – x)^2 e^x is:

i need solution

(a) Minimum value at x = –1; maximum value x = 1
(b) Minimum value at x = 1; maximum value x = –2 (c)
Minimum value at x = 1; maximum value x = –1
(d) None of these

To find the maximum or minimum values of a function, we need to take the derivative of the function and solve for where the derivative is equal to zero. In this case, we need to find the derivative of the function f(x) = (1 – x)^2 e^x.

Let's start by finding the derivative of f(x). We'll use the product rule here:

f'(x) = [(1 – x)^2]' e^x + (1 – x)^2 (e^x)'

The derivative of (1 – x)^2 is found by applying the chain rule:

[(1 – x)^2]'
= 2(1 – x)(-1)
= -2(1 – x)

The derivative of e^x is simply e^x.

So, the derivative of f(x) is:

f'(x) = -2(1 – x) e^x + (1 – x)^2 e^x

Next, we set the derivative equal to zero and solve for x:

f'(x) = 0
-2(1 – x) e^x + (1 – x)^2 e^x = 0

To simplify, let's factor out e^x:

e^x [-2(1 – x) + (1 – x)^2] = 0

Now, we have two cases to consider:

Case 1: e^x = 0
But e^x can never be equal to zero. So, this case is not possible.

Case 2: -2(1 – x) + (1 – x)^2 = 0
Expanding the equation, we get:
-2 + 2x + 1 - 2x + x^2 = 0
x^2 - 1 = 0

Factoring:
(x - 1)(x + 1) = 0

So, we have two possible values for x: x = 1 and x = -1.

To determine whether these values correspond to a maximum or minimum, we can use the second derivative test. However, to keep it simple, we can also plug in these values into the original function and compare the values.

Plugging in x = 1:
f(1) = (1 - 1)^2 e^1 = 0

Plugging in x = -1:
f(-1) = (1 - (-1))^2 e^(-1) = 4/e

Comparing the values, we can see that f(1) is 0, while f(-1) is positive (4/e). Therefore, f(-1) represents the minimum value of the function, and f(1) represents the maximum value of the function.

So, the correct answer is (c) Minimum value at x = 1; maximum value x = –1.