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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)-3x^(2)+9x+40, where
x > 0.
a) Where is
C(x) decreasing?
b) Where is
C(x) increasing?

Suppose the total $\operatorname{cost} C(x)$ (in dollars) to manufacture a quantity $x$ of weed killer (in hundreds of liters) is given by the function $C(x)=x^{3}-3 x^{2}+9 x+40$ , where $x>0$ . $\newline$ a) Where is $\mathrm{C}(\mathrm{x})$ decreasing? $\newline$ b) Where is $\mathrm{C}(\mathrm{x})$ increasing?

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Write linear functions: word problems

Full solution

Q. Suppose the total

\operatorname{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}-3 x^{2}+9 x+40

, where

x>0

.

\newline

a) Where is

\mathrm{C}(\mathrm{x})

decreasing?

\newline

b) Where is

\mathrm{C}(\mathrm{x})

increasing?

Calculate Derivative of $C(x)$ : To find where $C(x)$ is increasing or decreasing, we need to calculate the derivative of $C(x)$ , which is $C'(x)$ .
Differentiate $C(x)$ : Differentiate $C(x)$ with respect to $x$ to get $C'(x)$ .
$C'(x) = \frac{d}{dx} [x^3 - 3x^2 + 9x + 40]$
$C'(x) = 3x^2 - 6x + 9$
Find Critical Points: Find the critical points by setting $C'(x)$ equal to zero and solving for $x$ . $\newline$ $0 = 3x^2 - 6x + 9$
Solve Quadratic Equation: Solve the quadratic equation $3x^2 - 6x + 9 = 0$ . This equation does not factor easily, so we can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Use Quadratic Formula: Plug the coefficients into the quadratic formula. $\newline$ $a = 3$ , $b = -6$ , $c = 9$ $\newline$ $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(9)}}{2(3)}$ $\newline$ $x = \frac{6 \pm \sqrt{36 - 108}}{6}$ $\newline$ $x = \frac{6 \pm \sqrt{-72}}{6}$
Check Discriminant: Since the discriminant $\sqrt{-72}$ is negative, there are no real solutions to the equation. This means there are no critical points where $C'(x)$ changes sign.
Analyze Critical Points: Without real critical points, $C'(x)$ does not change sign and is always positive because the leading coefficient of $C'(x)$ is positive.
Determine Function Behavior: Since $C'(x)$ is always positive, $C(x)$ is always increasing for $x > 0$ .

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