Answer :
To find the intervals on which the function [tex]\( f(x) = -2x^3 - 24x^2 + 54x - 19 \)[/tex] is increasing, we need to follow these steps:
1. Find the derivative of the function: The first step is to find [tex]\( f'(x) \)[/tex], the derivative of [tex]\( f(x) \)[/tex]. The derivative gives us the rate of change of the function.
The derivative of each term in [tex]\( f(x) \)[/tex] is calculated as follows:
- The derivative of [tex]\(-2x^3\)[/tex] is [tex]\(-6x^2\)[/tex].
- The derivative of [tex]\(-24x^2\)[/tex] is [tex]\(-48x\)[/tex].
- The derivative of [tex]\(54x\)[/tex] is [tex]\(54\)[/tex].
- The derivative of [tex]\(-19\)[/tex] is [tex]\(0\)[/tex] since it's a constant.
So, the first derivative is:
[tex]\[
f'(x) = -6x^2 - 48x + 54
\][/tex]
2. Find the critical points: Critical points occur where [tex]\( f'(x) = 0 \)[/tex] or where [tex]\( f'(x) \)[/tex] is undefined. Since [tex]\( f'(x) \)[/tex] is a polynomial, it's defined everywhere. We only need to set [tex]\( f'(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex].
[tex]\[
-6x^2 - 48x + 54 = 0
\][/tex]
Divide the entire equation by [tex]\(-6\)[/tex] to simplify:
[tex]\[
x^2 + 8x - 9 = 0
\][/tex]
Factor the quadratic equation:
[tex]\[
(x + 9)(x - 1) = 0
\][/tex]
So the critical points are [tex]\( x = -9 \)[/tex] and [tex]\( x = 1 \)[/tex].
3. Determine the sign of [tex]\( f'(x) \)[/tex] around the critical points: We need to test the intervals determined by these critical points: [tex]\((- \infty, -9)\)[/tex], [tex]\((-9, 1)\)[/tex], and [tex]\((1, \infty)\)[/tex].
- For the interval [tex]\((- \infty, -9)\)[/tex], choose a test point, say [tex]\( x = -10 \)[/tex]:
[tex]\[
f'(-10) = -6(-10)^2 - 48(-10) + 54 = -600 + 480 + 54 = -66
\][/tex]
Since [tex]\(-66\)[/tex] is negative, [tex]\( f(x) \)[/tex] is decreasing on [tex]\((- \infty, -9)\)[/tex].
- For the interval [tex]\((-9, 1)\)[/tex], choose a test point, say [tex]\( x = 0 \)[/tex]:
[tex]\[
f'(0) = -6(0)^2 - 48(0) + 54 = 54
\][/tex]
Since [tex]\(54\)[/tex] is positive, [tex]\( f(x) \)[/tex] is increasing on [tex]\((-9, 1)\)[/tex].
- For the interval [tex]\((1, \infty)\)[/tex], choose a test point, say [tex]\( x = 2 \)[/tex]:
[tex]\[
f'(2) = -6(2)^2 - 48(2) + 54 = -24 - 96 + 54 = -66
\][/tex]
Since [tex]\(-66\)[/tex] is negative, [tex]\( f(x) \)[/tex] is decreasing on [tex]\((1, \infty)\)[/tex].
4. Conclusion: The function [tex]\( f(x) \)[/tex] is increasing on the interval [tex]\((-9, 1)\)[/tex].
1. Find the derivative of the function: The first step is to find [tex]\( f'(x) \)[/tex], the derivative of [tex]\( f(x) \)[/tex]. The derivative gives us the rate of change of the function.
The derivative of each term in [tex]\( f(x) \)[/tex] is calculated as follows:
- The derivative of [tex]\(-2x^3\)[/tex] is [tex]\(-6x^2\)[/tex].
- The derivative of [tex]\(-24x^2\)[/tex] is [tex]\(-48x\)[/tex].
- The derivative of [tex]\(54x\)[/tex] is [tex]\(54\)[/tex].
- The derivative of [tex]\(-19\)[/tex] is [tex]\(0\)[/tex] since it's a constant.
So, the first derivative is:
[tex]\[
f'(x) = -6x^2 - 48x + 54
\][/tex]
2. Find the critical points: Critical points occur where [tex]\( f'(x) = 0 \)[/tex] or where [tex]\( f'(x) \)[/tex] is undefined. Since [tex]\( f'(x) \)[/tex] is a polynomial, it's defined everywhere. We only need to set [tex]\( f'(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex].
[tex]\[
-6x^2 - 48x + 54 = 0
\][/tex]
Divide the entire equation by [tex]\(-6\)[/tex] to simplify:
[tex]\[
x^2 + 8x - 9 = 0
\][/tex]
Factor the quadratic equation:
[tex]\[
(x + 9)(x - 1) = 0
\][/tex]
So the critical points are [tex]\( x = -9 \)[/tex] and [tex]\( x = 1 \)[/tex].
3. Determine the sign of [tex]\( f'(x) \)[/tex] around the critical points: We need to test the intervals determined by these critical points: [tex]\((- \infty, -9)\)[/tex], [tex]\((-9, 1)\)[/tex], and [tex]\((1, \infty)\)[/tex].
- For the interval [tex]\((- \infty, -9)\)[/tex], choose a test point, say [tex]\( x = -10 \)[/tex]:
[tex]\[
f'(-10) = -6(-10)^2 - 48(-10) + 54 = -600 + 480 + 54 = -66
\][/tex]
Since [tex]\(-66\)[/tex] is negative, [tex]\( f(x) \)[/tex] is decreasing on [tex]\((- \infty, -9)\)[/tex].
- For the interval [tex]\((-9, 1)\)[/tex], choose a test point, say [tex]\( x = 0 \)[/tex]:
[tex]\[
f'(0) = -6(0)^2 - 48(0) + 54 = 54
\][/tex]
Since [tex]\(54\)[/tex] is positive, [tex]\( f(x) \)[/tex] is increasing on [tex]\((-9, 1)\)[/tex].
- For the interval [tex]\((1, \infty)\)[/tex], choose a test point, say [tex]\( x = 2 \)[/tex]:
[tex]\[
f'(2) = -6(2)^2 - 48(2) + 54 = -24 - 96 + 54 = -66
\][/tex]
Since [tex]\(-66\)[/tex] is negative, [tex]\( f(x) \)[/tex] is decreasing on [tex]\((1, \infty)\)[/tex].
4. Conclusion: The function [tex]\( f(x) \)[/tex] is increasing on the interval [tex]\((-9, 1)\)[/tex].
