instructions: show all work clearly. answers without supporting work will receive little or no credit.\n1…

instructions: show all work clearly. answers without supporting work will receive little or no credit.\n1. (20 points) consider the function\n$f(x)=2x^{3}-9x^{2}-24x + 5$.\n(a) (4 points) find the derivative $f(x)$.\n(b) (4 points) find all critical points of $f(x)$. show your algebra.\n(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.\n\\begin{tabular}{|l|l|l|}\n\\hline\ninterval for $x$ & sign of $f(x)$ ( & behavior of $f(x)$ \\\n\\hline\n$(-\\infty,\\quad)$ & & \\\n\\hline\n$(\\quad,\\quad)$ & & \\\n\\hline\n$(\\quad,\\infty)$ & & \\\n\\hline\n\\end{tabular}\n(d) (6 points) identify all relative extrema (local maxima and minima). write both the $x$-values and corresponding $f(x)$ values $(x,f(x))$.
Answer
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)) |
|---|---|---|
| ((-\infty,-1)) | (+) | Increasing |
| ((-1,4)) | (-) | Decreasing |
| ((4,\infty)) | (+) | Increasing |
(d) Relative maximum at ((-1,18)) and relative minimum at ((4,-107))