Answer :
To determine the intervals where the function [tex]\( f(x) = -3x^3 + 18x^2 + 45x + 19 \)[/tex] is decreasing, we need to find where its derivative, [tex]\( f'(x) \)[/tex], is negative. Here’s a step-by-step solution:
1. Find the Derivative:
The first step is to find the derivative [tex]\( f'(x) \)[/tex] of the function. For [tex]\( f(x) = -3x^3 + 18x^2 + 45x + 19 \)[/tex], the derivative is:
[tex]\[
f'(x) = \frac{d}{dx}(-3x^3 + 18x^2 + 45x + 19) = -9x^2 + 36x + 45.
\][/tex]
2. Find Critical Points:
Critical points occur where [tex]\( f'(x) = 0 \)[/tex] or where the derivative does not exist. Since this is a polynomial, [tex]\( f'(x) \)[/tex] exists everywhere, so we only need to solve [tex]\( -9x^2 + 36x + 45 = 0 \)[/tex].
3. Solve the Quadratic Equation:
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] for the equation [tex]\( -9x^2 + 36x + 45 = 0 \)[/tex], we find the roots:
[tex]\[
x = \frac{-36 \pm \sqrt{36^2 - 4(-9)(45)}}{2(-9)}
\][/tex]
Solving this yields the critical points [tex]\( x = -1 \)[/tex] and [tex]\( x = 5 \)[/tex].
4. Analyze the Intervals:
Use the critical points to divide the number line into intervals: [tex]\( (-\infty, -1) \)[/tex], [tex]\( (-1, 5) \)[/tex], and [tex]\( (5, \infty) \)[/tex].
5. Test the Intervals:
Pick test points in each interval to evaluate the sign of [tex]\( f'(x) \)[/tex]:
- For [tex]\( (-\infty, -1) \)[/tex], choose a test point like [tex]\( x = -2 \)[/tex]. Calculate [tex]\( f'(-2) \)[/tex] and check its sign.
- For [tex]\( (-1, 5) \)[/tex], choose a test point like [tex]\( x = 0 \)[/tex]. Calculate [tex]\( f'(0) \)[/tex] and check its sign.
- For [tex]\( (5, \infty) \)[/tex], choose a test point like [tex]\( x = 6 \)[/tex]. Calculate [tex]\( f'(6) \)[/tex] and check its sign.
6. Determine Decreasing Intervals:
You determine that:
- [tex]\( f'(x) < 0 \)[/tex] for [tex]\( (-\infty, -1) \)[/tex] and [tex]\( (5, \infty) \)[/tex].
Thus, the intervals where the function [tex]\( f(x) \)[/tex] is decreasing are [tex]\( (-\infty, -1) \)[/tex] and [tex]\( (5, \infty) \)[/tex].
1. Find the Derivative:
The first step is to find the derivative [tex]\( f'(x) \)[/tex] of the function. For [tex]\( f(x) = -3x^3 + 18x^2 + 45x + 19 \)[/tex], the derivative is:
[tex]\[
f'(x) = \frac{d}{dx}(-3x^3 + 18x^2 + 45x + 19) = -9x^2 + 36x + 45.
\][/tex]
2. Find Critical Points:
Critical points occur where [tex]\( f'(x) = 0 \)[/tex] or where the derivative does not exist. Since this is a polynomial, [tex]\( f'(x) \)[/tex] exists everywhere, so we only need to solve [tex]\( -9x^2 + 36x + 45 = 0 \)[/tex].
3. Solve the Quadratic Equation:
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] for the equation [tex]\( -9x^2 + 36x + 45 = 0 \)[/tex], we find the roots:
[tex]\[
x = \frac{-36 \pm \sqrt{36^2 - 4(-9)(45)}}{2(-9)}
\][/tex]
Solving this yields the critical points [tex]\( x = -1 \)[/tex] and [tex]\( x = 5 \)[/tex].
4. Analyze the Intervals:
Use the critical points to divide the number line into intervals: [tex]\( (-\infty, -1) \)[/tex], [tex]\( (-1, 5) \)[/tex], and [tex]\( (5, \infty) \)[/tex].
5. Test the Intervals:
Pick test points in each interval to evaluate the sign of [tex]\( f'(x) \)[/tex]:
- For [tex]\( (-\infty, -1) \)[/tex], choose a test point like [tex]\( x = -2 \)[/tex]. Calculate [tex]\( f'(-2) \)[/tex] and check its sign.
- For [tex]\( (-1, 5) \)[/tex], choose a test point like [tex]\( x = 0 \)[/tex]. Calculate [tex]\( f'(0) \)[/tex] and check its sign.
- For [tex]\( (5, \infty) \)[/tex], choose a test point like [tex]\( x = 6 \)[/tex]. Calculate [tex]\( f'(6) \)[/tex] and check its sign.
6. Determine Decreasing Intervals:
You determine that:
- [tex]\( f'(x) < 0 \)[/tex] for [tex]\( (-\infty, -1) \)[/tex] and [tex]\( (5, \infty) \)[/tex].
Thus, the intervals where the function [tex]\( f(x) \)[/tex] is decreasing are [tex]\( (-\infty, -1) \)[/tex] and [tex]\( (5, \infty) \)[/tex].
