Answer :
Let's solve the given problem step-by-step:
Consider the function [tex]\( f(x) = \frac{x^4 - 32 x^2}{9} \)[/tex].
### (a) First Derivative [tex]\( f'(x) \)[/tex]
To find the first derivative of [tex]\( f(x) \)[/tex], we will use the rules of differentiation.
[tex]\[ f'(x) = \frac{d}{dx}\left( \frac{x^4 - 32x^2}{9} \right) \][/tex]
Applying the power rule, we get:
[tex]\[ f'(x) = \frac{1}{9} \left( 4x^3 - 64x \right) = \frac{4x^3 - 64x}{9} \][/tex]
So, the first derivative is:
[tex]\[ f'(x) = \frac{4x^3 - 64x}{9} \][/tex]
### (b) Second Derivative [tex]\( f''(x) \)[/tex]
Next, we need to find the second derivative of [tex]\( f(x) \)[/tex].
[tex]\[ f''(x) = \frac{d}{dx}\left( \frac{4x^3 - 64x}{9} \right) \][/tex]
Applying the power rule again, we get:
[tex]\[ f''(x) = \frac{1}{9} \left( 12x^2 - 64 \right) = \frac{12x^2 - 64}{9} = \frac{4x^2 - 64}{3} \][/tex]
So, the second derivative is:
[tex]\[ f''(x) = \frac{4x^2 - 64}{3} \][/tex]
### (c) Interval of Increasing
For the function to be increasing, the first derivative [tex]\( f'(x) \)[/tex] must be greater than 0. We solve for [tex]\( f'(x) > 0 \)[/tex]:
However, the solution reveals that there are no intervals where [tex]\( f'(x) > 0 \)[/tex]. Thus, the function does not have any intervals of increasing.
So, the interval of increasing is:
[tex]\[ \text{interval of increasing} = \emptyset \][/tex]
### (d) Interval of Decreasing
For the function to be decreasing, the first derivative [tex]\( f'(x) \)[/tex] must be less than 0. We solve for [tex]\( f'(x) < 0 \)[/tex]:
Similarly, there are no intervals where [tex]\( f'(x) < 0 \)[/tex]. Thus, the function does not have any intervals of decreasing.
So, the interval of decreasing is:
[tex]\[ \text{interval of decreasing} = \emptyset \][/tex]
### (e) Interval of Concave Downward
For the function to be concave downward, the second derivative [tex]\( f''(x) \)[/tex] must be less than 0. We solve for [tex]\( f''(x) < 0 \)[/tex]:
The solution shows that there are no intervals where [tex]\( f''(x) < 0 \)[/tex]. Thus, the function does not have any intervals of concave downward.
So, the interval of concave downward is:
[tex]\[ \text{interval of concave downward} = \emptyset \][/tex]
### (f) Interval of Concave Upward
For the function to be concave upward, the second derivative [tex]\( f''(x) \)[/tex] must be greater than 0. We solve for [tex]\( f''(x) > 0 \)[/tex]:
The solution shows that there are no intervals where [tex]\( f''(x) > 0 \)[/tex]. Thus, the function does not have any intervals of concave upward.
So, the interval of concave upward is:
[tex]\[ \text{interval of concave upward} = \emptyset \][/tex]
Thus, after careful consideration, we conclude the following intervals:
(a) [tex]\( f'(x) = \frac{4x^3 - 64x}{9} \)[/tex]
(b) [tex]\( f''(x) = \frac{4x^2 - 64}{3} \)[/tex]
(c) Interval of increasing: [tex]\( \emptyset \)[/tex]
(d) Interval of decreasing: [tex]\( \emptyset \)[/tex]
(e) Interval of concave downward: [tex]\( \emptyset \)[/tex]
(f) Interval of concave upward: [tex]\( \emptyset \)[/tex]
Consider the function [tex]\( f(x) = \frac{x^4 - 32 x^2}{9} \)[/tex].
### (a) First Derivative [tex]\( f'(x) \)[/tex]
To find the first derivative of [tex]\( f(x) \)[/tex], we will use the rules of differentiation.
[tex]\[ f'(x) = \frac{d}{dx}\left( \frac{x^4 - 32x^2}{9} \right) \][/tex]
Applying the power rule, we get:
[tex]\[ f'(x) = \frac{1}{9} \left( 4x^3 - 64x \right) = \frac{4x^3 - 64x}{9} \][/tex]
So, the first derivative is:
[tex]\[ f'(x) = \frac{4x^3 - 64x}{9} \][/tex]
### (b) Second Derivative [tex]\( f''(x) \)[/tex]
Next, we need to find the second derivative of [tex]\( f(x) \)[/tex].
[tex]\[ f''(x) = \frac{d}{dx}\left( \frac{4x^3 - 64x}{9} \right) \][/tex]
Applying the power rule again, we get:
[tex]\[ f''(x) = \frac{1}{9} \left( 12x^2 - 64 \right) = \frac{12x^2 - 64}{9} = \frac{4x^2 - 64}{3} \][/tex]
So, the second derivative is:
[tex]\[ f''(x) = \frac{4x^2 - 64}{3} \][/tex]
### (c) Interval of Increasing
For the function to be increasing, the first derivative [tex]\( f'(x) \)[/tex] must be greater than 0. We solve for [tex]\( f'(x) > 0 \)[/tex]:
However, the solution reveals that there are no intervals where [tex]\( f'(x) > 0 \)[/tex]. Thus, the function does not have any intervals of increasing.
So, the interval of increasing is:
[tex]\[ \text{interval of increasing} = \emptyset \][/tex]
### (d) Interval of Decreasing
For the function to be decreasing, the first derivative [tex]\( f'(x) \)[/tex] must be less than 0. We solve for [tex]\( f'(x) < 0 \)[/tex]:
Similarly, there are no intervals where [tex]\( f'(x) < 0 \)[/tex]. Thus, the function does not have any intervals of decreasing.
So, the interval of decreasing is:
[tex]\[ \text{interval of decreasing} = \emptyset \][/tex]
### (e) Interval of Concave Downward
For the function to be concave downward, the second derivative [tex]\( f''(x) \)[/tex] must be less than 0. We solve for [tex]\( f''(x) < 0 \)[/tex]:
The solution shows that there are no intervals where [tex]\( f''(x) < 0 \)[/tex]. Thus, the function does not have any intervals of concave downward.
So, the interval of concave downward is:
[tex]\[ \text{interval of concave downward} = \emptyset \][/tex]
### (f) Interval of Concave Upward
For the function to be concave upward, the second derivative [tex]\( f''(x) \)[/tex] must be greater than 0. We solve for [tex]\( f''(x) > 0 \)[/tex]:
The solution shows that there are no intervals where [tex]\( f''(x) > 0 \)[/tex]. Thus, the function does not have any intervals of concave upward.
So, the interval of concave upward is:
[tex]\[ \text{interval of concave upward} = \emptyset \][/tex]
Thus, after careful consideration, we conclude the following intervals:
(a) [tex]\( f'(x) = \frac{4x^3 - 64x}{9} \)[/tex]
(b) [tex]\( f''(x) = \frac{4x^2 - 64}{3} \)[/tex]
(c) Interval of increasing: [tex]\( \emptyset \)[/tex]
(d) Interval of decreasing: [tex]\( \emptyset \)[/tex]
(e) Interval of concave downward: [tex]\( \emptyset \)[/tex]
(f) Interval of concave upward: [tex]\( \emptyset \)[/tex]
