Answer :
Sure, let's analyze the function [tex]\( f(x) = \frac{x^4 - 32x^2}{9} \)[/tex] step by step, using calculus:
### (a) First Derivative
To find the first derivative of [tex]\( f(x) \)[/tex], we use the power rule:
[tex]\[ f'(x) = \frac{d}{dx}\left( \frac{x^4 - 32x^2}{9} \right) \][/tex]
[tex]\[ f'(x) = \frac{4x^3 - 64x}{9} \][/tex]
### (b) Second Derivative
To find the second derivative of [tex]\( f(x) \)[/tex], we differentiate [tex]\( f'(x) \)[/tex]:
[tex]\[ f''(x) = \frac{d}{dx}\left( \frac{4x^3 - 64x}{9} \right) \][/tex]
[tex]\[ f''(x) = \frac{4x^2 - 64}{3} \][/tex]
### (c) Interval of Increasing
To find where the function is increasing, we solve [tex]\( f'(x) > 0 \)[/tex]:
[tex]\[ \frac{4x^3 - 64x}{9} > 0 \][/tex]
Factoring the numerator:
[tex]\[ \frac{4x(x^2 - 16)}{9} > 0 \][/tex]
Further factoring:
[tex]\[ \frac{4x(x - 4)(x + 4)}{9} > 0 \][/tex]
The critical points are [tex]\( x = -4, 0, 4 \)[/tex]. Using test intervals around these points, we find:
- The function is increasing on the intervals [tex]\( (-4, 0) \cup (4, \infty) \)[/tex].
### (d) Interval of Decreasing
To find where the function is decreasing, we solve [tex]\( f'(x) < 0 \)[/tex]:
[tex]\[ \frac{4x(x - 4)(x + 4)}{9} < 0 \][/tex]
Testing the intervals around the critical points [tex]\( x = -4, 0, 4 \)[/tex], we determine:
- The function is decreasing on the intervals [tex]\( (-\infty, -4) \cup (0, 4) \)[/tex].
### (e) Interval of Concave Downward
To find where the function is concave downward, we solve [tex]\( f''(x) < 0 \)[/tex]:
[tex]\[ \frac{4x^2 - 64}{3} < 0 \][/tex]
Simplifying:
[tex]\[ \frac{4(x^2 - 16)}{3} < 0 \][/tex]
Factor further:
[tex]\[ \frac{4(x - 4 \sqrt{3}/3)(x + 4 \sqrt{3}/3)}{3} < 0 \][/tex]
The critical points are [tex]\( x = \pm 4 \sqrt{3}/3 \)[/tex]. Testing the intervals around these points, we find:
- The function is concave downward on the interval [tex]\( -4 \sqrt{3}/3 < x < 4 \sqrt{3}/3 \)[/tex].
### (f) Interval of Concave Upward
To find where the function is concave upward, we solve [tex]\( f''(x) > 0 \)[/tex]:
[tex]\[ \frac{4x^2 - 64}{3} > 0 \][/tex]
Simplifying:
[tex]\[ \frac{4(x^2 - 16)}{3} > 0 \][/tex]
Factor further:
[tex]\[ \frac{4(x - 4 \sqrt{3}/3)(x + 4 \sqrt{3}/3)}{3} > 0 \][/tex]
Testing the intervals around the critical points [tex]\( x = \pm 4 \sqrt{3}/3 \)[/tex], we find:
- The function is concave upward on the intervals [tex]\( (-\infty, -4 \sqrt{3}/3) \cup (4 \sqrt{3}/3, \infty) \)[/tex].
To summarize, the intervals for each behavior of the function are given as:
(c) Interval of Increasing: [tex]\( (-4, 0) \cup (4, \infty) \)[/tex]
(d) Interval of Decreasing: [tex]\( (-\infty, -4) \cup (0, 4) \)[/tex]
(e) Interval of Concave Downward: [tex]\( -\infty < x < -4 \sqrt{3}/3 \cup 4 \sqrt{3}/3 < x < \infty \)[/tex]
(f) Interval of Concave Upward: [tex]\( (-\infty, -4 \sqrt{3}/3) \cup (4 \sqrt{3}/3, \infty) \)[/tex]
These intervals help us understand where the function is increasing, decreasing, concave upward, and concave downward.
### (a) First Derivative
To find the first derivative of [tex]\( f(x) \)[/tex], we use the power rule:
[tex]\[ f'(x) = \frac{d}{dx}\left( \frac{x^4 - 32x^2}{9} \right) \][/tex]
[tex]\[ f'(x) = \frac{4x^3 - 64x}{9} \][/tex]
### (b) Second Derivative
To find the second derivative of [tex]\( f(x) \)[/tex], we differentiate [tex]\( f'(x) \)[/tex]:
[tex]\[ f''(x) = \frac{d}{dx}\left( \frac{4x^3 - 64x}{9} \right) \][/tex]
[tex]\[ f''(x) = \frac{4x^2 - 64}{3} \][/tex]
### (c) Interval of Increasing
To find where the function is increasing, we solve [tex]\( f'(x) > 0 \)[/tex]:
[tex]\[ \frac{4x^3 - 64x}{9} > 0 \][/tex]
Factoring the numerator:
[tex]\[ \frac{4x(x^2 - 16)}{9} > 0 \][/tex]
Further factoring:
[tex]\[ \frac{4x(x - 4)(x + 4)}{9} > 0 \][/tex]
The critical points are [tex]\( x = -4, 0, 4 \)[/tex]. Using test intervals around these points, we find:
- The function is increasing on the intervals [tex]\( (-4, 0) \cup (4, \infty) \)[/tex].
### (d) Interval of Decreasing
To find where the function is decreasing, we solve [tex]\( f'(x) < 0 \)[/tex]:
[tex]\[ \frac{4x(x - 4)(x + 4)}{9} < 0 \][/tex]
Testing the intervals around the critical points [tex]\( x = -4, 0, 4 \)[/tex], we determine:
- The function is decreasing on the intervals [tex]\( (-\infty, -4) \cup (0, 4) \)[/tex].
### (e) Interval of Concave Downward
To find where the function is concave downward, we solve [tex]\( f''(x) < 0 \)[/tex]:
[tex]\[ \frac{4x^2 - 64}{3} < 0 \][/tex]
Simplifying:
[tex]\[ \frac{4(x^2 - 16)}{3} < 0 \][/tex]
Factor further:
[tex]\[ \frac{4(x - 4 \sqrt{3}/3)(x + 4 \sqrt{3}/3)}{3} < 0 \][/tex]
The critical points are [tex]\( x = \pm 4 \sqrt{3}/3 \)[/tex]. Testing the intervals around these points, we find:
- The function is concave downward on the interval [tex]\( -4 \sqrt{3}/3 < x < 4 \sqrt{3}/3 \)[/tex].
### (f) Interval of Concave Upward
To find where the function is concave upward, we solve [tex]\( f''(x) > 0 \)[/tex]:
[tex]\[ \frac{4x^2 - 64}{3} > 0 \][/tex]
Simplifying:
[tex]\[ \frac{4(x^2 - 16)}{3} > 0 \][/tex]
Factor further:
[tex]\[ \frac{4(x - 4 \sqrt{3}/3)(x + 4 \sqrt{3}/3)}{3} > 0 \][/tex]
Testing the intervals around the critical points [tex]\( x = \pm 4 \sqrt{3}/3 \)[/tex], we find:
- The function is concave upward on the intervals [tex]\( (-\infty, -4 \sqrt{3}/3) \cup (4 \sqrt{3}/3, \infty) \)[/tex].
To summarize, the intervals for each behavior of the function are given as:
(c) Interval of Increasing: [tex]\( (-4, 0) \cup (4, \infty) \)[/tex]
(d) Interval of Decreasing: [tex]\( (-\infty, -4) \cup (0, 4) \)[/tex]
(e) Interval of Concave Downward: [tex]\( -\infty < x < -4 \sqrt{3}/3 \cup 4 \sqrt{3}/3 < x < \infty \)[/tex]
(f) Interval of Concave Upward: [tex]\( (-\infty, -4 \sqrt{3}/3) \cup (4 \sqrt{3}/3, \infty) \)[/tex]
These intervals help us understand where the function is increasing, decreasing, concave upward, and concave downward.
