Answer :
To find the intervals on which the function [tex]\( f(x) = -3x^3 + 18x^2 + 45x + 19 \)[/tex] is decreasing, we need to follow a few steps:
1. Find the Derivative:
The first step is to find the derivative of [tex]\( f(x) \)[/tex]. The derivative, [tex]\( f'(x) \)[/tex], will help us determine where the function is increasing or decreasing.
[tex]\[
f'(x) = \frac{d}{dx}(-3x^3 + 18x^2 + 45x + 19) = -9x^2 + 36x + 45
\][/tex]
2. Determine Critical Points:
Set the derivative equal to zero to find the critical points.
[tex]\[
-9x^2 + 36x + 45 = 0
\][/tex]
This is a quadratic equation, which we can solve using the quadratic formula, [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].
For the given quadratic:
- [tex]\( a = -9 \)[/tex]
- [tex]\( b = 36 \)[/tex]
- [tex]\( c = 45 \)[/tex]
Plug these into the formula:
[tex]\[
x = \frac{-36 \pm \sqrt{36^2 - 4(-9)(45)}}{2(-9)}
\][/tex]
[tex]\[
x = \frac{-36 \pm \sqrt{1296 + 1620}}{-18}
\][/tex]
[tex]\[
x = \frac{-36 \pm \sqrt{2916}}{-18}
\][/tex]
[tex]\[
x = \frac{-36 \pm 54}{-18}
\][/tex]
Solving further, we find:
[tex]\[
x = \frac{-36 + 54}{-18} = -1
\quad \text{and} \quad
x = \frac{-36 - 54}{-18} = 5
\][/tex]
Thus, the critical points are [tex]\( x = -1 \)[/tex] and [tex]\( x = 5 \)[/tex].
3. Test Intervals Around Critical Points:
We need to test intervals around these critical points to determine where [tex]\( f'(x) \)[/tex] is negative, indicating that [tex]\( f(x) \)[/tex] is decreasing.
Choose test points in the intervals [tex]\( (-\infty, -1) \)[/tex], [tex]\( (-1, 5) \)[/tex], and [tex]\( (5, \infty) \)[/tex]:
- For the interval [tex]\( (-\infty, -1) \)[/tex], choose [tex]\( x = -2 \)[/tex]:
[tex]\[
f'(-2) = -9(-2)^2 + 36(-2) + 45 = -36 - 72 + 45 = -63 \quad (\text{negative})
\][/tex]
- For the interval [tex]\( (-1, 5) \)[/tex], choose [tex]\( x = 0 \)[/tex]:
[tex]\[
f'(0) = -9(0)^2 + 36(0) + 45 = 45 \quad (\text{positive})
\][/tex]
- For the interval [tex]\( (5, \infty) \)[/tex], choose [tex]\( x = 6 \)[/tex]:
[tex]\[
f'(6) = -9(6)^2 + 36(6) + 45 = -324 + 216 + 45 = -63 \quad (\text{negative})
\][/tex]
4. Identify Decreasing Intervals:
From our tests, [tex]\( f'(x) \)[/tex] is negative in the intervals [tex]\( (-\infty, -1) \)[/tex] and [tex]\( (5, \infty) \)[/tex]. Therefore, [tex]\( f(x) \)[/tex] is decreasing on these intervals.
In conclusion, the function [tex]\( f(x) = -3x^3 + 18x^2 + 45x + 19 \)[/tex] is decreasing on the intervals [tex]\((- \infty, -1)\)[/tex] and [tex]\((5, \infty)\)[/tex].
1. Find the Derivative:
The first step is to find the derivative of [tex]\( f(x) \)[/tex]. The derivative, [tex]\( f'(x) \)[/tex], will help us determine where the function is increasing or decreasing.
[tex]\[
f'(x) = \frac{d}{dx}(-3x^3 + 18x^2 + 45x + 19) = -9x^2 + 36x + 45
\][/tex]
2. Determine Critical Points:
Set the derivative equal to zero to find the critical points.
[tex]\[
-9x^2 + 36x + 45 = 0
\][/tex]
This is a quadratic equation, which we can solve using the quadratic formula, [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].
For the given quadratic:
- [tex]\( a = -9 \)[/tex]
- [tex]\( b = 36 \)[/tex]
- [tex]\( c = 45 \)[/tex]
Plug these into the formula:
[tex]\[
x = \frac{-36 \pm \sqrt{36^2 - 4(-9)(45)}}{2(-9)}
\][/tex]
[tex]\[
x = \frac{-36 \pm \sqrt{1296 + 1620}}{-18}
\][/tex]
[tex]\[
x = \frac{-36 \pm \sqrt{2916}}{-18}
\][/tex]
[tex]\[
x = \frac{-36 \pm 54}{-18}
\][/tex]
Solving further, we find:
[tex]\[
x = \frac{-36 + 54}{-18} = -1
\quad \text{and} \quad
x = \frac{-36 - 54}{-18} = 5
\][/tex]
Thus, the critical points are [tex]\( x = -1 \)[/tex] and [tex]\( x = 5 \)[/tex].
3. Test Intervals Around Critical Points:
We need to test intervals around these critical points to determine where [tex]\( f'(x) \)[/tex] is negative, indicating that [tex]\( f(x) \)[/tex] is decreasing.
Choose test points in the intervals [tex]\( (-\infty, -1) \)[/tex], [tex]\( (-1, 5) \)[/tex], and [tex]\( (5, \infty) \)[/tex]:
- For the interval [tex]\( (-\infty, -1) \)[/tex], choose [tex]\( x = -2 \)[/tex]:
[tex]\[
f'(-2) = -9(-2)^2 + 36(-2) + 45 = -36 - 72 + 45 = -63 \quad (\text{negative})
\][/tex]
- For the interval [tex]\( (-1, 5) \)[/tex], choose [tex]\( x = 0 \)[/tex]:
[tex]\[
f'(0) = -9(0)^2 + 36(0) + 45 = 45 \quad (\text{positive})
\][/tex]
- For the interval [tex]\( (5, \infty) \)[/tex], choose [tex]\( x = 6 \)[/tex]:
[tex]\[
f'(6) = -9(6)^2 + 36(6) + 45 = -324 + 216 + 45 = -63 \quad (\text{negative})
\][/tex]
4. Identify Decreasing Intervals:
From our tests, [tex]\( f'(x) \)[/tex] is negative in the intervals [tex]\( (-\infty, -1) \)[/tex] and [tex]\( (5, \infty) \)[/tex]. Therefore, [tex]\( f(x) \)[/tex] is decreasing on these intervals.
In conclusion, the function [tex]\( f(x) = -3x^3 + 18x^2 + 45x + 19 \)[/tex] is decreasing on the intervals [tex]\((- \infty, -1)\)[/tex] and [tex]\((5, \infty)\)[/tex].
