Question

On what intervals is the function f(x)=x⁸ - 8x⁷ both decreasing and concave up? interval(s) =

(Give your answer as an interval or a list of intervals, e.g. (-intinity,8] or (1,5),(7,10).

Answer 1

The function f(x) = x^8 - 8x^7 is both decreasing and concave up on the interval (0, 6).

Step-by-step explanation:

To determine on which intervals the function f(x) = x⁸ - 8x⁷ is both decreasing and concave up, we need to find the intervals where f'(x) is negative (decreasing) and f''(x) is positive (concave up).

Step 1: Find the first derivative f'(x)

The first derivative of the function is f'(x) = 8x⁷ - 56x⁶.

Step 2: Find the second derivative f''(x)

The second derivative of the function is f''(x) = 56x⁶ - 336x⁵.

Step 3: Find intervals where f'(x) < 0

Set the first derivative less than zero and solve for x to find where the function is decreasing. We find that f'(x) < 0 for x in (0, 7).

Step 4: Find intervals where f''(x) > 0

Set the second derivative greater than zero and solve for x to find where the function is concave up. We find that f''(x) > 0 for x in (0, 6).

Step 5: Find the overlap of these intervals

The function is both decreasing and concave up where the intervals from Step 3 and Step 4 overlap, which is (0, 6).

Answer 2

The function f(x) = x⁸ - 8x⁷ is decreasing and concave up over the interval:

x ∈ (-∞, 1)

Step-by-step explanation:

Analyzing the first derivative:

f'(x) = 8x⁶(x-1)

f'(x) = 0 for x = 0, 1 (critical points)

Intervals and signs of f'(x):

| Interval | x-value | f'(x) sign | Monotonicity |

|---|---|---|---|

| x < 0 | -1 | + | Increasing |

| 0 < x < 1 | 0.5 | - | Decreasing |

| x > 1 | 2 | + | Increasing |

Analyzing the second derivative:

f''(x) = 48x⁵(3x-2)

Intervals and signs of f''(x):

| Interval | x-value | f''(x) sign | Concavity |

|---|---|---|---|

| x < 0 | -1 | + | Concave Up |

| 0 < x < 2/3 | 0.1 | - | Concave Down |

| x > 2/3 | 1 | + | Concave Up |

Combining the information:

The function is decreasing and concave up only in the interval where f'(x) is negative and f''(x) is positive, which is:

x ∈ (-∞, 1)

Therefore, the function satisfies the given conditions over the interval (-∞, 1).