Find all relative extrema of the function. Use the second derivative test where applicable. (If an answer does not exist, enter DNE.)f(x) = x4 − 8x3 + 1

To find the relative extrema of the function f(x) = x^4 - 8x^3 + 1, we need to follow these steps:

1. Find the derivative of the function: f'(x) = 4x^3 - 24x^2.

2. Set the derivative equal to zero and solve for x to find critical points: 4x^3 - 24x^2 = 0. Factoring out 4x^2 gives 4x^2(x - 6) = 0. Setting each factor equal to zero gives x = 0 and x = 6.

3. Find the second derivative of the function: f''(x) = 12x^2 - 48x.

4. Substitute the critical points into the second derivative to determine whether each is a maximum, minimum, or point of inflection.

   • f''(0) = 12(0)^2 - 48(0) = 0, which does not give us any information.

   • f''(6) = 12(6)^2 - 48(6) = 0, which also does not give us any information.

5. Since the second derivative test failed, we need to use the first derivative test. We pick a number to the left and right of each critical point and evaluate the first derivative at these points.

   • For x = 0, pick -1 and 1. f'(-1) = 4(-1)^3 - 24(-1)^2 = -4 - 24 = -28, which is negative, and f'(1) = 4(1)^3 - 24(1)^2 = 4 - 24 = -20, which is also negative. Since the derivative does not change sign, x = 0 is not a relative extremum.

   • For x = 6, pick 5 and 7. f'(5) = 4(5)^3 - 24(5)^2 = 500 - 600 = -100, which is negative, and f'(7) = 4(7)^3 - 24(7)^2 = 1372 - 1176 = 196, which is positive. Since the derivative changes sign from negative to positive, x = 6 is a relative minimum.

So, the function f(x) = x^4 - 8x^3 + 1 has a relative minimum at x = 6. There are no relative maximums.