Answer :
To determine where the function [tex]\( f(x) = 2x^3 - 24x^2 + 90x + 19 \)[/tex] is decreasing, we must first find its critical points and then analyze the intervals around these points.
### Step 1: Find the Derivative
The first step is to calculate the derivative of [tex]\( f(x) \)[/tex]. The derivative, denoted as [tex]\( f'(x) \)[/tex], represents the rate of change of the function.
[tex]\[ f'(x) = \frac{d}{dx}(2x^3 - 24x^2 + 90x + 19) \][/tex]
Applying the power rule of differentiation, we get:
[tex]\[ f'(x) = 6x^2 - 48x + 90 \][/tex]
### Step 2: Find Critical Points
Critical points occur where the derivative is zero or undefined. Since the derivative [tex]\( f'(x) = 6x^2 - 48x + 90 \)[/tex] is a polynomial, it's defined everywhere. So, we set the derivative equal to zero to find the critical points:
[tex]\[ 6x^2 - 48x + 90 = 0 \][/tex]
Solving this quadratic equation will give the critical points. For this equation, the critical points are [tex]\( x = 3 \)[/tex] and [tex]\( x = 5 \)[/tex].
### Step 3: Determine Intervals of Increase and Decrease
To determine where the function is decreasing, we need to check the sign of [tex]\( f'(x) \)[/tex] in the intervals determined by the critical points. The critical points split the number line into the following intervals: [tex]\((-∞, 3)\)[/tex], [tex]\((3, 5)\)[/tex], and [tex]\((5, ∞)\)[/tex].
- Interval [tex]\((-∞, 3)\)[/tex]: Pick a test value such as [tex]\( x = 0 \)[/tex].
- Substitute into [tex]\( f'(x) \)[/tex]: [tex]\( f'(0) = 6(0)^2 - 48(0) + 90 = 90 \)[/tex].
- Since 90 is positive, [tex]\( f(x) \)[/tex] is increasing on [tex]\((-∞, 3)\)[/tex].
- Interval [tex]\((3, 5)\)[/tex]: Pick a test value such as [tex]\( x = 4 \)[/tex].
- Substitute into [tex]\( f'(x) \)[/tex]: [tex]\( f'(4) = 6(4)^2 - 48(4) + 90 = 6(16) - 192 + 90 = 96 - 192 + 90 = -6 \)[/tex].
- Since [tex]\(-6\)[/tex] is negative, [tex]\( f(x) \)[/tex] is decreasing on [tex]\((3, 5)\)[/tex].
- Interval [tex]\((5, ∞)\)[/tex]: Pick a test value such as [tex]\( x = 6 \)[/tex].
- Substitute into [tex]\( f'(x) \)[/tex]: [tex]\( f'(6) = 6(6)^2 - 48(6) + 90 = 216 - 288 + 90 = 18 \)[/tex].
- Since 18 is positive, [tex]\( f(x) \)[/tex] is increasing on [tex]\((5, ∞)\)[/tex].
### Conclusion
The function [tex]\( f(x) = 2x^3 - 24x^2 + 90x + 19 \)[/tex] is decreasing on the interval [tex]\((3, 5)\)[/tex].
### Step 1: Find the Derivative
The first step is to calculate the derivative of [tex]\( f(x) \)[/tex]. The derivative, denoted as [tex]\( f'(x) \)[/tex], represents the rate of change of the function.
[tex]\[ f'(x) = \frac{d}{dx}(2x^3 - 24x^2 + 90x + 19) \][/tex]
Applying the power rule of differentiation, we get:
[tex]\[ f'(x) = 6x^2 - 48x + 90 \][/tex]
### Step 2: Find Critical Points
Critical points occur where the derivative is zero or undefined. Since the derivative [tex]\( f'(x) = 6x^2 - 48x + 90 \)[/tex] is a polynomial, it's defined everywhere. So, we set the derivative equal to zero to find the critical points:
[tex]\[ 6x^2 - 48x + 90 = 0 \][/tex]
Solving this quadratic equation will give the critical points. For this equation, the critical points are [tex]\( x = 3 \)[/tex] and [tex]\( x = 5 \)[/tex].
### Step 3: Determine Intervals of Increase and Decrease
To determine where the function is decreasing, we need to check the sign of [tex]\( f'(x) \)[/tex] in the intervals determined by the critical points. The critical points split the number line into the following intervals: [tex]\((-∞, 3)\)[/tex], [tex]\((3, 5)\)[/tex], and [tex]\((5, ∞)\)[/tex].
- Interval [tex]\((-∞, 3)\)[/tex]: Pick a test value such as [tex]\( x = 0 \)[/tex].
- Substitute into [tex]\( f'(x) \)[/tex]: [tex]\( f'(0) = 6(0)^2 - 48(0) + 90 = 90 \)[/tex].
- Since 90 is positive, [tex]\( f(x) \)[/tex] is increasing on [tex]\((-∞, 3)\)[/tex].
- Interval [tex]\((3, 5)\)[/tex]: Pick a test value such as [tex]\( x = 4 \)[/tex].
- Substitute into [tex]\( f'(x) \)[/tex]: [tex]\( f'(4) = 6(4)^2 - 48(4) + 90 = 6(16) - 192 + 90 = 96 - 192 + 90 = -6 \)[/tex].
- Since [tex]\(-6\)[/tex] is negative, [tex]\( f(x) \)[/tex] is decreasing on [tex]\((3, 5)\)[/tex].
- Interval [tex]\((5, ∞)\)[/tex]: Pick a test value such as [tex]\( x = 6 \)[/tex].
- Substitute into [tex]\( f'(x) \)[/tex]: [tex]\( f'(6) = 6(6)^2 - 48(6) + 90 = 216 - 288 + 90 = 18 \)[/tex].
- Since 18 is positive, [tex]\( f(x) \)[/tex] is increasing on [tex]\((5, ∞)\)[/tex].
### Conclusion
The function [tex]\( f(x) = 2x^3 - 24x^2 + 90x + 19 \)[/tex] is decreasing on the interval [tex]\((3, 5)\)[/tex].
