for x is an element of [-11,11]the function defined by f(x)=x^7(x+3)^6. where does the function achieve its minimum?

To find where the function achieves its minimum, we need to find the critical points of the function and check the endpoints of the given interval.

1. Find the derivative of the function:
f'(x) = 7x^6(x+3)^6 + 6x^7(x+3)^5

2. Set the derivative equal to zero to find the critical points:
7x^6(x+3)^6 + 6x^7(x+3)^5 = 0

Since the derivative consists of two terms being multiplied, we can set each term equal to zero and solve for x:

7x^6 = 0 --> x = 0 (as x raised to an even power will always be positive)

(x+3)^6 = 0 --> x = -3

3. Now, we need to check the value of the function at these critical points and the endpoints of the interval [-11, 11]:

- At x = -11: f(-11) = (-11)^7(-11+3)^6
- At x = -3 (critical point): f(-3) = (-3)^7(-3+3)^6
- At x = 0 (critical point): f(0) = (0)^7(0+3)^6
- At x = 11: f(11) = (11)^7(11+3)^6

4. Calculate the value of the function at each point.

- At x = -11: f(-11) = (-11)^7(-11+3)^6 ≈ -7.72 × 10^10
- At x = -3 (critical point): f(-3) = (-3)^7(-3+3)^6 = 0
- At x = 0 (critical point): f(0) = (0)^7(0+3)^6 = 0
- At x = 11: f(11) = (11)^7(11+3)^6 ≈ 2.12 × 10^11

5. Compare all the calculated values.

- f(-11) ≈ -7.72 × 10^10
- f(-3) = 0
- f(0) = 0
- f(11) ≈ 2.12 × 10^11

By comparing these values, we can see that the function achieves its minimum at x = -3 and x = 0. Both of these critical points result in a function value of zero, which means the function is at its minimum at these points.

Therefore, the function achieves its minimum at x = -3 and x = 0 within the given interval [-11,11].