Answer :
To find the intervals on which the function [tex]\( f(x) = 2x^3 - 24x^2 + 90x + 19 \)[/tex] is decreasing, we can follow these steps:
1. Find the derivative [tex]\( f'(x) \)[/tex]:
The derivative of the function [tex]\( f(x) \)[/tex] is given by differentiating each term:
[tex]\[
f'(x) = \frac{d}{dx}(2x^3 - 24x^2 + 90x + 19) = 6x^2 - 48x + 90
\][/tex]
2. Find critical points:
Critical points occur where the derivative is equal to zero or undefined. In this case, we solve [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[
6x^2 - 48x + 90 = 0
\][/tex]
Dividing the entire equation by 6 to simplify:
[tex]\[
x^2 - 8x + 15 = 0
\][/tex]
This quadratic equation can be factored as:
[tex]\[
(x - 3)(x - 5) = 0
\][/tex]
Therefore, the critical points are [tex]\( x = 3 \)[/tex] and [tex]\( x = 5 \)[/tex].
3. Determine the intervals:
The critical points divide the x-axis into intervals: [tex]\((- \infty, 3)\)[/tex], [tex]\((3, 5)\)[/tex], and [tex]\((5, \infty)\)[/tex].
4. Test the sign of [tex]\( f'(x) \)[/tex] on each interval:
We choose a test point from each interval to determine if the function is increasing or decreasing in that interval.
- For [tex]\((- \infty, 3)\)[/tex], choose [tex]\( x = 0 \)[/tex]:
[tex]\[
f'(0) = 6(0)^2 - 48(0) + 90 = 90 > 0 \][/tex]
The function is increasing on [tex]\((- \infty, 3)\)[/tex].
- For [tex]\((3, 5)\)[/tex], choose [tex]\( x = 4 \)[/tex]:
[tex]\[
f'(4) = 6(4)^2 - 48(4) + 90 = 96 - 192 + 90 = -6 < 0 \][/tex]
The function is decreasing on [tex]\((3, 5)\)[/tex].
- For [tex]\((5, \infty)\)[/tex], choose [tex]\( x = 6 \)[/tex]:
[tex]\[
f'(6) = 6(6)^2 - 48(6) + 90 = 216 - 288 + 90 = 18 > 0 \][/tex]
The function is increasing on [tex]\((5, \infty)\)[/tex].
5. State the intervals where the function is decreasing:
From our tests, the function is decreasing on the interval [tex]\((3, 5)\)[/tex].
Therefore, the interval on which [tex]\( f(x) \)[/tex] is decreasing is [tex]\((3, 5)\)[/tex].
1. Find the derivative [tex]\( f'(x) \)[/tex]:
The derivative of the function [tex]\( f(x) \)[/tex] is given by differentiating each term:
[tex]\[
f'(x) = \frac{d}{dx}(2x^3 - 24x^2 + 90x + 19) = 6x^2 - 48x + 90
\][/tex]
2. Find critical points:
Critical points occur where the derivative is equal to zero or undefined. In this case, we solve [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[
6x^2 - 48x + 90 = 0
\][/tex]
Dividing the entire equation by 6 to simplify:
[tex]\[
x^2 - 8x + 15 = 0
\][/tex]
This quadratic equation can be factored as:
[tex]\[
(x - 3)(x - 5) = 0
\][/tex]
Therefore, the critical points are [tex]\( x = 3 \)[/tex] and [tex]\( x = 5 \)[/tex].
3. Determine the intervals:
The critical points divide the x-axis into intervals: [tex]\((- \infty, 3)\)[/tex], [tex]\((3, 5)\)[/tex], and [tex]\((5, \infty)\)[/tex].
4. Test the sign of [tex]\( f'(x) \)[/tex] on each interval:
We choose a test point from each interval to determine if the function is increasing or decreasing in that interval.
- For [tex]\((- \infty, 3)\)[/tex], choose [tex]\( x = 0 \)[/tex]:
[tex]\[
f'(0) = 6(0)^2 - 48(0) + 90 = 90 > 0 \][/tex]
The function is increasing on [tex]\((- \infty, 3)\)[/tex].
- For [tex]\((3, 5)\)[/tex], choose [tex]\( x = 4 \)[/tex]:
[tex]\[
f'(4) = 6(4)^2 - 48(4) + 90 = 96 - 192 + 90 = -6 < 0 \][/tex]
The function is decreasing on [tex]\((3, 5)\)[/tex].
- For [tex]\((5, \infty)\)[/tex], choose [tex]\( x = 6 \)[/tex]:
[tex]\[
f'(6) = 6(6)^2 - 48(6) + 90 = 216 - 288 + 90 = 18 > 0 \][/tex]
The function is increasing on [tex]\((5, \infty)\)[/tex].
5. State the intervals where the function is decreasing:
From our tests, the function is decreasing on the interval [tex]\((3, 5)\)[/tex].
Therefore, the interval on which [tex]\( f(x) \)[/tex] is decreasing is [tex]\((3, 5)\)[/tex].
