Answer :
To solve the problem step-by-step, we'll follow these steps:
1. Find the first derivative of [tex]\( f(x) \)[/tex].
2. Find the second derivative of [tex]\( f(x) \)[/tex].
3. Determine the intervals where [tex]\( f(x) \)[/tex] is increasing or decreasing.
4. Determine the intervals where [tex]\( f(x) \)[/tex] is concave upward or downward.
Given:
[tex]\[ f(x) = x^3 + 3x^2 - 45x + 11 \][/tex]
### (a) Find the first derivative of [tex]\( f(x) \)[/tex]
Using the power rule [tex]\(\left(\frac{d}{dx} x^n = nx^{n-1}\right)\)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}\left(x^3 + 3x^2 - 45x + 11\right) = 3x^2 + 6x - 45 \][/tex]
So,
[tex]\[ f'(x) = 3x^2 + 6x - 45 \][/tex]
### (b) Find the second derivative of [tex]\( f(x) \)[/tex]
Again using the power rule:
[tex]\[ f''(x) = \frac{d}{dx}\left(3x^2 + 6x - 45\right) = 6x + 6 \][/tex]
So,
[tex]\[ f''(x) = 6x + 6 \][/tex]
### (c) Determine the intervals where [tex]\( f(x) \)[/tex] is increasing
First, find the critical points by setting [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ 3x^2 + 6x - 45 = 0 \][/tex]
Divide the equation by 3:
[tex]\[ x^2 + 2x - 15 = 0 \][/tex]
Factor:
[tex]\[ (x + 5)(x - 3) = 0 \][/tex]
So, the critical points are:
[tex]\[ x = -5 \quad \text{and} \quad x = 3 \][/tex]
To determine the intervals of increase and decrease, test the sign of [tex]\( f'(x) \)[/tex] in the intervals determined by the critical points [tex]\((- \infty, -5)\)[/tex], [tex]\((-5, 3)\)[/tex], and [tex]\((3, \infty)\)[/tex]:
- For [tex]\( x \in (-\infty, -5) \)[/tex], choose [tex]\( x = -6 \)[/tex]:
[tex]\[ f'(-6) = 3(-6)^2 + 6(-6) - 45 = 108 - 36 - 45 = 27 \quad (\text{positive}) \][/tex]
- For [tex]\( x \in (-5, 3) \)[/tex], choose [tex]\( x = 0 \)[/tex]:
[tex]\[ f'(0) = 3(0)^2 + 6(0) - 45 = -45 \quad (\text{negative}) \][/tex]
- For [tex]\( x \in (3, \infty) \)[/tex], choose [tex]\( x = 4 \)[/tex]:
[tex]\[ f'(4) = 3(4)^2 + 6(4) - 45 = 48 + 24 - 45 = 27 \quad (\text{positive}) \][/tex]
So, [tex]\( f(x) \)[/tex] is increasing on:
[tex]\[ (-\infty, -5] \cup [3, \infty) \][/tex]
### (d) Determine the intervals where [tex]\( f(x) \)[/tex] is decreasing
Using the results from (c), [tex]\( f(x) \)[/tex] is decreasing on:
[tex]\[ [-5, 3] \][/tex]
### (e) Determine the intervals where [tex]\( f(x) \)[/tex] is concave downward
Find the inflection points by setting [tex]\( f''(x) = 0 \)[/tex]:
[tex]\[ 6x + 6 = 0 \][/tex]
[tex]\[ x = -1 \][/tex]
Test the sign of [tex]\( f''(x) \)[/tex] in the intervals [tex]\((- \infty, -1)\)[/tex] and [tex]\((-1, \infty)\)[/tex]:
- For [tex]\( x \in (-\infty, -1) \)[/tex], choose [tex]\( x = -2 \)[/tex]:
[tex]\[ f''(-2) = 6(-2) + 6 = -12 + 6 = -6 \quad (\text{negative}) \][/tex]
- For [tex]\( x \in (-1, \infty) \)[/tex], choose [tex]\( x = 0 \)[/tex]:
[tex]\[ f''(0) = 6(0) + 6 = 6 \quad (\text{positive}) \][/tex]
So, [tex]\( f(x) \)[/tex] is concave downward on:
[tex]\[ (-\infty, -1] \][/tex]
### (f) Determine the intervals where [tex]\( f(x) \)[/tex] is concave upward
Using the results from (e), [tex]\( f(x) \)[/tex] is concave upward on:
[tex]\[ [-1, \infty) \][/tex]
### Summary:
- [tex]\((a)\)[/tex] [tex]\( f'(x) = 3x^2 + 6x - 45 \)[/tex]
- [tex]\((b)\)[/tex] [tex]\( f''(x) = 6x + 6 \)[/tex]
- [tex]\((c)\)[/tex] Interval of increasing: [tex]\( (-\infty, -5] \cup [3, \infty) \)[/tex]
- [tex]\((d)\)[/tex] Interval of decreasing: [tex]\( [-5, 3] \)[/tex]
- [tex]\((e)\)[/tex] Interval of concave downward: [tex]\( (-\infty, -1] \)[/tex]
- [tex]\((f)\)[/tex] Interval of concave upward: [tex]\( [-1, \infty) \)[/tex]
1. Find the first derivative of [tex]\( f(x) \)[/tex].
2. Find the second derivative of [tex]\( f(x) \)[/tex].
3. Determine the intervals where [tex]\( f(x) \)[/tex] is increasing or decreasing.
4. Determine the intervals where [tex]\( f(x) \)[/tex] is concave upward or downward.
Given:
[tex]\[ f(x) = x^3 + 3x^2 - 45x + 11 \][/tex]
### (a) Find the first derivative of [tex]\( f(x) \)[/tex]
Using the power rule [tex]\(\left(\frac{d}{dx} x^n = nx^{n-1}\right)\)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}\left(x^3 + 3x^2 - 45x + 11\right) = 3x^2 + 6x - 45 \][/tex]
So,
[tex]\[ f'(x) = 3x^2 + 6x - 45 \][/tex]
### (b) Find the second derivative of [tex]\( f(x) \)[/tex]
Again using the power rule:
[tex]\[ f''(x) = \frac{d}{dx}\left(3x^2 + 6x - 45\right) = 6x + 6 \][/tex]
So,
[tex]\[ f''(x) = 6x + 6 \][/tex]
### (c) Determine the intervals where [tex]\( f(x) \)[/tex] is increasing
First, find the critical points by setting [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ 3x^2 + 6x - 45 = 0 \][/tex]
Divide the equation by 3:
[tex]\[ x^2 + 2x - 15 = 0 \][/tex]
Factor:
[tex]\[ (x + 5)(x - 3) = 0 \][/tex]
So, the critical points are:
[tex]\[ x = -5 \quad \text{and} \quad x = 3 \][/tex]
To determine the intervals of increase and decrease, test the sign of [tex]\( f'(x) \)[/tex] in the intervals determined by the critical points [tex]\((- \infty, -5)\)[/tex], [tex]\((-5, 3)\)[/tex], and [tex]\((3, \infty)\)[/tex]:
- For [tex]\( x \in (-\infty, -5) \)[/tex], choose [tex]\( x = -6 \)[/tex]:
[tex]\[ f'(-6) = 3(-6)^2 + 6(-6) - 45 = 108 - 36 - 45 = 27 \quad (\text{positive}) \][/tex]
- For [tex]\( x \in (-5, 3) \)[/tex], choose [tex]\( x = 0 \)[/tex]:
[tex]\[ f'(0) = 3(0)^2 + 6(0) - 45 = -45 \quad (\text{negative}) \][/tex]
- For [tex]\( x \in (3, \infty) \)[/tex], choose [tex]\( x = 4 \)[/tex]:
[tex]\[ f'(4) = 3(4)^2 + 6(4) - 45 = 48 + 24 - 45 = 27 \quad (\text{positive}) \][/tex]
So, [tex]\( f(x) \)[/tex] is increasing on:
[tex]\[ (-\infty, -5] \cup [3, \infty) \][/tex]
### (d) Determine the intervals where [tex]\( f(x) \)[/tex] is decreasing
Using the results from (c), [tex]\( f(x) \)[/tex] is decreasing on:
[tex]\[ [-5, 3] \][/tex]
### (e) Determine the intervals where [tex]\( f(x) \)[/tex] is concave downward
Find the inflection points by setting [tex]\( f''(x) = 0 \)[/tex]:
[tex]\[ 6x + 6 = 0 \][/tex]
[tex]\[ x = -1 \][/tex]
Test the sign of [tex]\( f''(x) \)[/tex] in the intervals [tex]\((- \infty, -1)\)[/tex] and [tex]\((-1, \infty)\)[/tex]:
- For [tex]\( x \in (-\infty, -1) \)[/tex], choose [tex]\( x = -2 \)[/tex]:
[tex]\[ f''(-2) = 6(-2) + 6 = -12 + 6 = -6 \quad (\text{negative}) \][/tex]
- For [tex]\( x \in (-1, \infty) \)[/tex], choose [tex]\( x = 0 \)[/tex]:
[tex]\[ f''(0) = 6(0) + 6 = 6 \quad (\text{positive}) \][/tex]
So, [tex]\( f(x) \)[/tex] is concave downward on:
[tex]\[ (-\infty, -1] \][/tex]
### (f) Determine the intervals where [tex]\( f(x) \)[/tex] is concave upward
Using the results from (e), [tex]\( f(x) \)[/tex] is concave upward on:
[tex]\[ [-1, \infty) \][/tex]
### Summary:
- [tex]\((a)\)[/tex] [tex]\( f'(x) = 3x^2 + 6x - 45 \)[/tex]
- [tex]\((b)\)[/tex] [tex]\( f''(x) = 6x + 6 \)[/tex]
- [tex]\((c)\)[/tex] Interval of increasing: [tex]\( (-\infty, -5] \cup [3, \infty) \)[/tex]
- [tex]\((d)\)[/tex] Interval of decreasing: [tex]\( [-5, 3] \)[/tex]
- [tex]\((e)\)[/tex] Interval of concave downward: [tex]\( (-\infty, -1] \)[/tex]
- [tex]\((f)\)[/tex] Interval of concave upward: [tex]\( [-1, \infty) \)[/tex]
