Let's go through each part of the question step-by-step.
Given the function [tex]\( f(x) = x^3 + 3x^2 - 45x + 2 \)[/tex], we answer the questions as follows:
### (a) First Derivative
To find the first derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(x^3 + 3x^2 - 45x + 2) \][/tex]
Using the power rule [tex]\( \left(\frac{d}{dx} x^n = nx^{n-1}\right) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(3x^2) + \frac{d}{dx}(-45x) + \frac{d}{dx}(2) \][/tex]
[tex]\[ f'(x) = 3x^2 + 6x - 45 \][/tex]
### (b) Second Derivative
To find the second derivative [tex]\( f''(x) \)[/tex]:
[tex]\[ f''(x) = \frac{d}{dx}(3x^2 + 6x - 45) \][/tex]
Using the power rule:
[tex]\[ f''(x) = \frac{d}{dx}(3x^2) + \frac{d}{dx}(6x) + \frac{d}{dx}(-45) \][/tex]
[tex]\[ f''(x) = 6x + 6 \][/tex]
### (c) Intervals of Increasing
To find the intervals where [tex]\( f \)[/tex] is increasing, find critical points by setting [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ 3x^2 + 6x - 45 = 0 \][/tex]
Solve this quadratic equation for [tex]\( x \)[/tex]:
[tex]\[ x^2 + 2x - 15 = 0 \][/tex]
[tex]\[ (x + 5)(x - 3) = 0 \][/tex]
[tex]\[ x = -5, \, 3 \][/tex]
Now, analyze the sign of [tex]\( f'(x) \)[/tex] around these critical points. The intervals are divided by [tex]\( x = -5 \)[/tex] and [tex]\( x = 3 \)[/tex]:
- For [tex]\( x < -5 \)[/tex]: [tex]\( f'(x) > 0 \)[/tex]
- For [tex]\( -5 < x < 3 \)[/tex]: [tex]\( f'(x) < 0 \)[/tex]
- For [tex]\( x > 3 \)[/tex]: [tex]\( f'(x) > 0 \)[/tex]
Thus, the function [tex]\( f \)[/tex] is increasing on the intervals:
Interval of increasing = [tex]\([-5, \infty)\)[/tex]
### (d) Intervals of Decreasing
From the above analysis:
Interval of decreasing = [tex]\((-\infty, -5]\)[/tex]
### (e) Intervals of Concave Downward
To find the intervals of concavity, solve [tex]\( f''(x) = 0 \)[/tex]:
[tex]\[ 6x + 6 = 0 \][/tex]
[tex]\[ x = -1 \][/tex]
Now, analyze the sign of [tex]\( f''(x) \)[/tex] around [tex]\( x = -1 \)[/tex]:
- For [tex]\( x < -1 \)[/tex]: [tex]\( f''(x) < 0 \)[/tex]
- For [tex]\( x > -1 \)[/tex]: [tex]\( f''(x) > 0 \)[/tex]
Thus, the function [tex]\( f \)[/tex] is concave downward on the interval:
Interval of downward concavity = [tex]\((-\infty, -1]\)[/tex]
### (f) Intervals of Concave Upward
From the above analysis:
Interval of upward concavity = [tex]\([-1, \infty)\)[/tex]
In summary:
(c) The interval where [tex]\( f \)[/tex] is increasing: [tex]\([-5, \infty)\)[/tex]
(d) The interval where [tex]\( f \)[/tex] is decreasing: [tex]\((-\infty, -5]\)[/tex]
(e) The interval where [tex]\( f \)[/tex] is concave downward: [tex]\((-\infty, -1]\)[/tex]
(f) The interval where [tex]\( f \)[/tex] is concave upward: [tex]\([-1, \infty)\)[/tex]
