Question

F(x)=x^(3)+6x^(2)-96x, determine all intervals on which f is both decreasing and concave down.

Answer 1

To find the intervals on which f(x) is both decreasing and concave down, we need to analyze the signs of its first and second derivatives. The function f(x) = x^3 + 6x^2 - 96x is decreasing on the intervals (-∞, -8) and (4, ∞), and it is concave down on the interval (-∞, -2).

Step-by-step explanation:

To determine the intervals on which a function is both decreasing and concave down, we need to analyze the signs of its first and second derivatives. Let's find the first and second derivatives off(x):

f'(x) = 3x^2 + 12x - 96

f''(x) = 6x + 12

To find the intervals where f(x) is decreasing, we need to find where f'(x) is negative. Setting f'(x) < 0:

3x^2 + 12x - 96 < 0

Solving the quadratic inequality, we find x < -8 or x > 4. So, f(x) is decreasing on the intervals (-∞, -8) and (4, ∞).

To find the intervals where f(x) is concave down, we need f''(x) < 0. Setting f''(x) < 0:

6x + 12 < 0

Solving the inequality, we find x < -2. Therefore, f(x) is concave down on the interval (-∞, -2).

Answer 2

There are no common intervals where the function is both decreasing and concave down. This means that within the domain of f(x) , there are no intervals that satisfy both conditions simultaneously.

To determine the intervals where the function
`\( f(x) = x^3 + 6x^2 - 96x \)`is both decreasing and concave down, we need to follow these steps:

1. Find the first derivative, f'(x) , to analyze where the function is increasing or decreasing.

2. Find the second derivative, f''(x) to determine the concavity of the function.

3. Analyze critical points and inflection points to establish the intervals where the function is decreasing and concave down.

Step 1: First Derivative f'(x)

`\[ f'(x) = (d)/(dx)(x^3 + 6x^2 - 96x) \]`

Step 2: Second Derivative f''(x)

`\[ f''(x) = (d^2)/(dx^2)(x^3 + 6x^2 - 96x) \]`

Step 3: Analyze Critical Points and Inflection Points

- Find critical points by solving f'(x) = 0 .

- Determine intervals of increase/decrease around these points.

- Find inflection points by solving f''(x) = 0

- Determine intervals of concavity around these points.

- Identify intervals where the function is both decreasing and concave down.

Let's start by calculating f'(x) and f''(x)

The first and second derivatives of the function are as follows:

1. First derivative
`\( f'(x) = 3x^2 + 12x - 96 \)`

2. Second derivative
`\( f''(x) = 6x + 12 \)`

Step 3: Analyze Critical Points and Inflection Points

Critical Points from f'(x)

We find the critical points by solving f'(x) = 0 .

Inflection Points (from f''(x) )

We find the inflection points by solving f''(x) = 0 .

Let's solve these equations to find the critical and inflection points.

The critical points and inflection points are as follows:

- Critical Points: x = -8 and x = 4

- Inflection Point: x = -2

Analyzing Intervals

For Decreasing Function

We need to check the sign of f'(x) around the critical points -8 and 4 If
`\( f'(x) < 0 \)` in an interval, the function is decreasing in that interval.

For Concave Down Function

We check the sign of f''(x) around the inflection point
`\( -2 \). If \( f''(x) < 0 \)`in an interval, the function is concave down in that interval.

Combining Both Conditions

We'll identify the intervals where the function is both decreasing and concave down by intersecting the intervals obtained from the above two conditions.

Let's perform these sign tests.

It seems there was an error in defining the intervals using `sympy`. Let me correct that and proceed with the sign tests for both the first and second derivatives.

It appears there was an oversight in my code. I forgot to import the symbol representing infinity from `sympy`. Let me correct this and proceed with the calculations.

Based on the calculations:

1. The intervals where the function
`\( f(x) = x^3 + 6x^2 - 96x \) is decreasing are \( x \in (-8, 4) \).`

2. The intervals where the function is concave down are
`\( x \in (-\infty, -2) \).`