Questions: Suppose the total cost C(x) (in dollars) to manufacture a quantity x of weed killer (in hundreds of liters) is given by the function C(x)=x^3-2x^2+3x+50, where x>0. a) Where is C(x) decreasing? b) Where is C(x) increasing? a) Select the correct choice below and fill in any answer boxes within your choice. A. The function is decreasing on the open interval(s) 1. (Type your answer in interval notation. Simplify your answer. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) B. The function is never decreasing.

$Suppose the total cost C(x) (in dollars) to manufacture a quantity x of weed killer (in hundreds of liters) is given by the function C(x)=x^3-2x^2+3x+50, where x>0. a) Where is C(x) decreasing? b) Where is C(x) increasing? a) Select the correct choice below and fill in any answer boxes within your choice. A. The function is decreasing on the open interval(s) 1. (Type your answer in interval notation. Simplify your answer. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) B. The function is never decreasing.$

Transcript text: Suppose the total cost $C(x)$ (in dollars) to manufacture a quantity $x$ of weed killer (in hundreds of liters) is given by the function $C(x)=x^{3}-2 x^{2}+3 x+50$, where $x>0$. a) Where is $C(x)$ decreasing? b) Where is $C(x)$ increasing? a) Select the correct choice below and fill in any answer boxes within your choice. A. The function is decreasing on the open interval(s) $\square$ 1. (Type your answer in interval notation. Simplify your answer. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) B. The function is never decreasing.

Solution

Solution Steps

To determine where the function $ C(x) = x^3 - 2x^2 + 3x + 50 $ is decreasing or increasing, we need to find its first derivative $ C'(x) $ and analyze its critical points. The function is decreasing where the first derivative is negative and increasing where the first derivative is positive.

Compute the first derivative $ C'(x) $.
Find the critical points by setting $ C'(x) = 0 $ and solving for $ x $.
Determine the intervals where $ C'(x) $ is positive or negative by testing points in each interval.

Step 1: Find the First Derivative of $ C(x) $

To determine where the function $ C(x) $ is increasing or decreasing, we first need to find its first derivative. The given function is: \[ C(x) = x^3 - 2x^2 + 3x + 50 \]

The first derivative of $ C(x) $ is: \[ C'(x) = \frac{d}{dx}(x^3 - 2x^2 + 3x + 50) \] \[ C'(x) = 3x^2 - 4x + 3 \]

Step 2: Find the Critical Points

To find the critical points, we set the first derivative equal to zero and solve for $ x $: \[ 3x^2 - 4x + 3 = 0 \]

This is a quadratic equation. We solve it using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where $ a = 3 $, $ b = -4 $, and $ c = 3 $.

\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} \] \[ x = \frac{4 \pm \sqrt{16 - 36}}{6} \] \[ x = \frac{4 \pm \sqrt{-20}}{6} \] \[ x = \frac{4 \pm 2i\sqrt{5}}{6} \] \[ x = \frac{2 \pm i\sqrt{5}}{3} \]

Since the solutions are complex, there are no real critical points.

Step 3: Determine the Behavior of $ C(x) $

Since there are no real critical points, we need to analyze the sign of $ C'(x) $ over the entire domain $ x > 0 $.

\[ C'(x) = 3x^2 - 4x + 3 \]

We can test the sign of $ C'(x) $ for $ x > 0 $. Let's choose a test point, say $ x = 1 $:

\[ C'(1) = 3(1)^2 - 4(1) + 3 \] \[ C'(1) = 3 - 4 + 3 \] \[ C'(1) = 2 \]

Since $ C'(1) > 0 $, $ C'(x) $ is positive for $ x > 0 $.