Question

instructions: show all work clearly. answers without supporting work will receive little or no credit.

1. (20 points) consider the function

$f(x)=2x^{3}-9x^{2}-24x + 5$.
(a) (4 points) find the derivative $f(x)$.
(b) (4 points) find all critical points of $f(x)$. show your algebra.
(c) (6 points) complete the following sign chart (or make one below) for $f(x)$, and identify intervals where $f(x)$ is increasing or decreasing.
\

$$\begin{tabular}{|l|l|l|} \\hline interval for $x$ & sign of $f(x)$ ( & behavior of $f(x)$ \\ \\hline $(-\\infty,\\quad)$ & & \\ \\hline $(\\quad,\\quad)$ & & \\ \\hline $(\\quad,\\infty)$ & & \\ \\hline \\end{tabular}$$

(d) (6 points) identify all relative extrema (local maxima and minima). write both the $x$-values and corresponding $f(x)$ values $(x,f(x))$.

Explanation:

Step1: Differentiate the function \(f(x)\)

Using the power rule \(\frac{d}{dx}(x^n)=nx^{n - 1}\), for \(f(x)=2x^{3}-9x^{2}-24x + 5\), we have:
\(f^{\prime}(x)=\frac{d}{dx}(2x^{3})-\frac{d}{dx}(9x^{2})-\frac{d}{dx}(24x)+\frac{d}{dx}(5)\)
\(f^{\prime}(x)=2\times3x^{2}-9\times2x-24\times1+0\)
\(f^{\prime}(x)=6x^{2}-18x - 24\)

Step2: Find the critical points

Set \(f^{\prime}(x) = 0\), so \(6x^{2}-18x - 24=0\). Divide through by \(6\): \(x^{2}-3x - 4=0\).
Factor the quadratic equation: \(x^{2}-3x - 4=(x + 1)(x - 4)=0\).
Using the zero - product property \(x+1 = 0\) or \(x - 4=0\), so \(x=-1\) or \(x = 4\)

Step3: Analyze the sign of \(f^{\prime}(x)\)

Choose test points:

• For the interval \((-\infty,-1)\), let \(x=-2\). Then \(f^{\prime}(-2)=6\times(-2)^{2}-18\times(-2)-24=6\times4 + 36-24=24 + 36-24=36>0\)
• For the interval \((-1,4)\), let \(x = 0\). Then \(f^{\prime}(0)=6\times0^{2}-18\times0-24=-24<0\)
• For the interval \((4,\infty)\), let \(x = 5\). Then \(f^{\prime}(5)=6\times5^{2}-18\times5-24=6\times25-90 - 24=150-90-24 = 36>0\)

The function \(f(x)\) is increasing when \(f^{\prime}(x)>0\) (on \((-\infty,-1)\cup(4,\infty)\)) and decreasing when \(f^{\prime}(x)<0\) (on \((-1,4)\))

Step4: Find the relative extrema

• For \(x=-1\): \(f(-1)=2\times(-1)^{3}-9\times(-1)^{2}-24\times(-1)+5=-2-9 + 24+5=18\)
• For \(x = 4\): \(f(4)=2\times4^{3}-9\times4^{2}-24\times4+5=2\times64-9\times16-96 + 5=128-144-96 + 5=-107\)

Answer:

(a) \(f^{\prime}(x)=6x^{2}-18x - 24\)

(b) Critical points at \(x=-1\) and \(x = 4\)

(c)

Interval for \(x\)	Sign of \(f^{\prime}(x)\)	Behavior of \(f(x)\)
\((-1,4)\)	\(-\)	Decreasing
\((4,\infty)\)	\(+\)	Increasing

(d) Relative maximum at \((-1,18)\) and relative minimum at \((4,-107)\)