Consider the function $ f(x) = 6x + 2x - 1 $.

For this function, there are four important intervals: $(-∞, A]$, $(A, B)$, $(B, C]$, and $[C, ∞)$, where $ A $ and $ C $ are the critical numbers, and the function is not defined at $ B $.

1. Find $ A $, $ B $, and $ C $.
2. For each of the following open intervals, tell whether $ f(x) $ is increasing or decreasing:
- $(-∞, A)$
- $(A, B)$
- $(B, C)$
- $(C, ∞)$

Note that this function has no inflection points, but we can still consider its concavity.

3. For each of the following intervals, tell whether $ f(x) $ is concave up or concave down:
- $(-∞, B)$
- $(B, ∞)$

Consider the function $ f(x) = 12x^5 + 60x^4 - 100x^3 + 1 $.

The function $ f(x) $ has inflection points at (reading from left to right) $ x = D, E, $ and $ F $.

4. For each of the following intervals, tell whether $ f(x) $ is concave up or concave down:
- $(-∞, D)$
- $(D, E)$
- $(E, F)$
- $(F, ∞)$

To find the critical numbers and points of discontinuity for the function f(x) = 6x + 2x^(-1), let's first find the derivative and identify where it is undefined.

Find the derivative of f(x):

f'(x) = 6 - 2x^(-2)

Identify points of discontinuity:

The function is not defined when x = 0 since x^(-1) is undefined at x = 0.

Find critical numbers:

To find the critical numbers, we set the derivative equal to zero and solve for x:

6 - 2x^(-2) = 0

2x^(-2) = 6

x^(-2) = 3/2

Taking the reciprocal of both sides:

x^2 = 2/3

x = ±√(2/3)

So, the critical numbers are A = -√(2/3) and C = √(2/3). The function is not defined at B = 0.

Now let's determine whether f(x) is increasing or decreasing and its concavity in each of the given intervals:

(−∞, A):

In this interval, f(x) is decreasing since f'(x) = 6 - 2x^(-2) > 0 for x < A.

(A, B):

In this interval, f(x) is increasing since f'(x) = 6 - 2x^(-2) > 0 for A < x < B.

(B, C):

The function is not defined in this interval.

(C, ∞):In this interval, f(x) is increasing since f'(x) = 6 - 2x^(-2) > 0 for x > C.

Next, let's consider the function f(x) = 12x^5 + 60x^4 - 100x^3 + 1:

To find the inflection points, we need to find where the concavity changes. This occurs when the second derivative is zero or undefined.

Find the second derivative of f(x):

f''(x) = 60x^4 + 240x^3 - 300x^2

Find the points of inflection:

Set f''(x) = 0 and solve for x:

60x^4 + 240x^3 - 300x^2 = 0

60x^2(x^2 + 4x - 5) = 0

x^2 + 4x - 5 = 0

Solving this quadratic equation, we find two solutions: x = D and x = E, where D < E.

Determine concavity:

(−∞, D):

In this interval, f(x) is concave up since f''(x) = 60x^4 + 240x^3 - 300x^2 > 0 for x < D.

(D, E):

In this interval, f(x) is concave down since f''(x) = 60x^4 + 240x^3 - 300x^2 < 0 for D < x < E.

(E, ∞):

In this interval, f(x) is concave up since f''(x) = 60x^4 + 240x^3 - 300x^2 > 0 for x > E.

Therefore, the concavity of f(x) is as follows:

(−∞, D): Concave up

(D, E): Concave down

(E, ∞): Concave up

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