hw 16 - first derivative test section 3.2: problem 5\n(1 point)\nconsider the function\n$f(x)=x^{3}-9x^{2}+15…

hw 16 - first derivative test section 3.2: problem 5\n(1 point)\nconsider the function\n$f(x)=x^{3}-9x^{2}+15x + 3$.\n1. $f$ is increasing on the intervals\n2. $f$ is decreasing on the intervals\n3. the relative maxima of $f$ occur at $x=$\n4. the relative minima of $f$ occur at $x=$\nnotes: in the last two, your answer should be a comma separated list of $x$ values or the word \none\.\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 6 attempts remaining.
Answer
Explanation:
Step1: Find the first derivative
The derivative of (f(x)=x^{3}-9x^{2}+15x + 3) is (f^\prime(x)=3x^{2}-18x + 15). Factor (f^\prime(x)): (f^\prime(x)=3(x^{2}-6x + 5)=3(x - 1)(x - 5)).
Step2: Find critical points
Set (f^\prime(x)=0), then (3(x - 1)(x - 5)=0). Solving for (x), we get (x = 1) and (x = 5).
Step3: Determine intervals of increase and decrease
Use test - points.
- For (x\lt1), let (x = 0). Then (f^\prime(0)=3(0 - 1)(0 - 5)=15\gt0). So (f(x)) is increasing on ((-\infty,1)).
- For (1\lt x\lt5), let (x = 2). Then (f^\prime(2)=3(2 - 1)(2 - 5)=-9\lt0). So (f(x)) is decreasing on ((1,5)).
- For (x\gt5), let (x = 6). Then (f^\prime(6)=3(6 - 1)(6 - 5)=15\gt0). So (f(x)) is increasing on ((5,\infty)).
Step4: Find relative maxima and minima
By the first - derivative test:
- Since (f(x)) changes from increasing to decreasing at (x = 1), (f(x)) has a relative maximum at (x = 1).
- Since (f(x)) changes from decreasing to increasing at (x = 5), (f(x)) has a relative minimum at (x = 5).
Answer:
- (f) is increasing on the intervals ((-\infty,1)\cup(5,\infty))
- (f) is decreasing on the intervals ((1,5))
- The relative maxima of (f) occur at (x = 1)
- The relative minima of (f) occur at (x = 5)