for the following function, a) give the coordinates of any critical points and classify each point as a…

for the following function, a) give the coordinates of any critical points and classify each point as a relative maximum, a relative minimum, or neither, b) identify intervals where the function is increasing or decreasing, c) give the coordinates of any points of inflection, d) identify intervals where the function is concave up or concave down, and e) sketch the graph\n$g(x)=x^{3}-12x^{2}+45x + 8$\n
Answer
Explanation:
Step1: Find the first derivative
The function is (g(x)=x^{3}-12x^{2}+45x + 8). Using the power rule ((x^{n})^\prime=nx^{n - 1}), we have (g^\prime(x)=3x^{2}-24x + 45). Factor out 3: (g^\prime(x)=3(x^{2}-8x + 15)=3(x - 3)(x - 5)). Set (g^\prime(x)=0), then (3(x - 3)(x - 5)=0). Solving for (x), we get (x = 3) and (x = 5). When (x = 3), (g(3)=3^{3}-12\times3^{2}+45\times3 + 8=27-108 + 135+8=62). When (x = 5), (g(5)=5^{3}-12\times5^{2}+45\times5 + 8=125-300 + 225+8=58). To classify the critical points, we use the second - derivative test. Find the second derivative: (g^{\prime\prime}(x)=6x-24). When (x = 3), (g^{\prime\prime}(3)=6\times3-24=-6<0), so ((3,62)) is a relative maximum. When (x = 5), (g^{\prime\prime}(5)=6\times5-24 = 6>0), so ((5,58)) is a relative minimum.
Step2: Determine the intervals of increase and decrease
Test intervals for (g^\prime(x)):
- For (x<3), let (x = 0), (g^\prime(0)=3(0 - 3)(0 - 5)=45>0). So (g(x)) is increasing on ((-\infty,3)).
- For (3<x<5), let (x = 4), (g^\prime(4)=3(4 - 3)(4 - 5)=-3<0). So (g(x)) is decreasing on ((3,5)).
- For (x>5), let (x = 6), (g^\prime(6)=3(6 - 3)(6 - 5)=9>0). So (g(x)) is increasing on ((5,\infty)).
Step3: Find the points of inflection
Set (g^{\prime\prime}(x)=0), (6x-24 = 0), solving for (x) gives (x = 4). When (x = 4), (g(4)=4^{3}-12\times4^{2}+45\times4 + 8=64-192+180 + 8=60). So the point of inflection is ((4,60)).
Step4: Determine the intervals of concavity
Test intervals for (g^{\prime\prime}(x)):
- For (x<4), let (x = 3), (g^{\prime\prime}(3)=6\times3-24=-6<0). So (g(x)) is concave down on ((-\infty,4)).
- For (x>4), let (x = 5), (g^{\prime\prime}(5)=6\times5-24 = 6>0). So (g(x)) is concave up on ((4,\infty)).
Answer:
a) Critical points: ((3,62)) (relative maximum), ((5,58)) (relative minimum). b) Increasing intervals: ((-\infty,3)\cup(5,\infty)); Decreasing interval: ((3,5)). c) Point of inflection: ((4,60)). d) Concave down interval: ((-\infty,4)); Concave up interval: ((4,\infty)).