2. for this problem, consider the function ( g(x)=\frac{x^{3}}{x + 2} ).\n(a) find the critical points of (…

2. for this problem, consider the function ( g(x)=\frac{x^{3}}{x + 2} ).\n(a) find the critical points of ( g(x) ).\n( \frac{x + 2(2 x)-x^{2}(1)}{(x + 2)^{2}} quad \frac{2 x^{2}+4 x-x^{2}}{(x + 2)^{2}} quad \frac{x^{2}+2 x}{(x + 2)^{2}} quad x(x + 4) )\n( x = 0,-4 )\ncritical points:\n(b) find the interval(s) on which ( g(x) ) is increasing and the interval(s) on which ( g(x) ) is decreasing.\nnote: keep in mind that ( g(x) ) has a vertical asymptote.\n(c) find ( g^{prime prime}(x) ).\n( g^{prime prime}(x)=)\n(d) use the second derivative test to classify the critical points you found in (b) as local minima,\nlocal maxima, or neither.

2. for this problem, consider the function ( g(x)=\frac{x^{3}}{x + 2} ).\n(a) find the critical points of ( g(x) ).\n( \frac{x + 2(2 x)-x^{2}(1)}{(x + 2)^{2}} quad \frac{2 x^{2}+4 x-x^{2}}{(x + 2)^{2}} quad \frac{x^{2}+2 x}{(x + 2)^{2}} quad x(x + 4) )\n( x = 0,-4 )\ncritical points:\n(b) find the interval(s) on which ( g(x) ) is increasing and the interval(s) on which ( g(x) ) is decreasing.\nnote: keep in mind that ( g(x) ) has a vertical asymptote.\n(c) find ( g^{prime prime}(x) ).\n( g^{prime prime}(x)=)\n(d) use the second derivative test to classify the critical points you found in (b) as local minima,\nlocal maxima, or neither.

Answer

Explanation:

Step1: Simplify the function

Given (g(x)=\frac{x^{3}}{x + 2}). Using the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = x^{3}), (u^\prime=3x^{2}), (v=x + 2), (v^\prime = 1). The first - derivative (g^\prime(x)=\frac{3x^{2}(x + 2)-x^{3}\times1}{(x + 2)^{2}}=\frac{3x^{3}+6x^{2}-x^{3}}{(x + 2)^{2}}=\frac{2x^{3}+6x^{2}}{(x + 2)^{2}}=\frac{2x^{2}(x + 3)}{(x + 2)^{2}}). Set (g^\prime(x)=0), then (2x^{2}(x + 3)=0) (since ((x + 2)^{2}\neq0) for critical points). Solving (2x^{2}(x + 3)=0) gives (x = 0) and (x=-3).

Step2: Find the second - derivative

Using the quotient rule again. Let (u = 2x^{3}+6x^{2}), (u^\prime=6x^{2}+12x), (v=(x + 2)^{2}), (v^\prime = 2(x + 2)). (g^{\prime\prime}(x)=\frac{(6x^{2}+12x)(x + 2)^{2}-(2x^{3}+6x^{2})\times2(x + 2)}{(x + 2)^{4}}) Factor out (2(x + 2)) from the numerator: (g^{\prime\prime}(x)=\frac{2(x + 2)[(3x^{2}+6x)(x + 2)-(2x^{3}+6x^{2})]}{(x + 2)^{4}}=\frac{2[(3x^{3}+6x^{2}+6x^{2}+12x)-(2x^{3}+6x^{2})]}{(x + 2)^{3}}) Simplify the numerator: (2(3x^{3}+6x^{2}+6x^{2}+12x - 2x^{3}-6x^{2})=2(x^{3}+6x^{2}+12x)) So (g^{\prime\prime}(x)=\frac{2x(x^{2}+6x + 12)}{(x + 2)^{3}})

Step3: Apply the second - derivative test

For (x=-3): (g^{\prime\prime}(-3)=\frac{2\times(-3)[(-3)^{2}+6\times(-3)+12]}{(-3 + 2)^{3}}=\frac{-6(9-18 + 12)}{-1}=\frac{-6\times3}{-1}=18>0) For (x = 0): (g^{\prime\prime}(0)=\frac{2\times0\times(0^{2}+6\times0 + 12)}{(0 + 2)^{3}}=0)

Answer:

  • (a) Critical points: (x=-3) and (x = 0)
  • (b) Increasing on ((-\infty,-3)\cup(-2,\infty)) (since (g^\prime(x)>0) when (x<-3) or (x>-2) and (x\neq - 2)), decreasing on ((-3,-2)) (since (g^\prime(x)<0) when (-3<x<-2))
  • (c) (g^{\prime\prime}(x)=\frac{2x(x^{2}+6x + 12)}{(x + 2)^{3}})
  • (d) (x=-3) is a local minimum (because (g^{\prime\prime}(-3)>0)), and the second - derivative test is inconclusive for (x = 0) (because (g^{\prime\prime}(0)=0))