Given:
[tex]C=\frac{x^4}{4}-\frac{4}{3}x^3-\frac{35}{2}x^2+150x[/tex]Let's solve for the following:
• (a). Find C'(x).
Here, we are to find the derivative of C(x).
Apply the sum rule:
[tex]\begin{gathered} C^{\prime}=\frac{d}{dx}(\frac{x^4}{4})+\frac{d}{dx}(-\frac{4}{3}x^3)+\frac{d}{dx}(-\frac{35}{2}x^2)+\frac{d}{dx}(150x) \\ \\ C^{\prime}=x^3-4x^2-35x+150 \end{gathered}[/tex]• (b). The critical numbers of C(x).
The critical numbers will be the points where the graph changes direction.
Using the derivative, let's solve for x.
[tex]x^3-4x^2-35x+150=0[/tex]Factor the left side using the rational root test:
[tex](x-5)(x^2+x-30)=0[/tex]Now factor using the AC method:
[tex]\begin{gathered} (x-5)((x-5)(x+6))=0 \\ \\ (x-5)(x-5)(x+6)=0 \end{gathered}[/tex]Equate each factor to zero and solve for x:
[tex]\begin{gathered} x-5=0 \\ \text{ Add 5 to both sides:} \\ x-5+5=0+5 \\ x=5 \\ \\ \\ x-5=0 \\ x-5+5=0+5 \\ x=5 \\ \\ \\ x+6=0 \\ x+6-6=0-6 \\ x=-6 \end{gathered}[/tex]Therefore, the critical numbers of C(x) are:
x = -6, 5
• (C). Increasing interval.
Use the critical points to find the increasing and decreasing intervals.
Using interval notation, the increasing interval is:
[tex](-6,5)\cup(5,\infty)[/tex]• D. Decreasing interval:
Using interval notation, the decreasing interval is:
[tex](-\infty,-6)[/tex]ANSWER:
(a). C'(x) = x³ - 4x² - 35x + 150
(b). x = -6, 5
(c). Increasing: (-6, 5) U (5,∞)
Decreasing: (-∞, -6)
