which of the following statements are true about the function graphed here? select all that apply.\nthe…

which of the following statements are true about the function graphed here? select all that apply.\nthe mathematical function is increasing for all x such\nthat 15 < x < 22.\nthe mathematical function is decreasing for all values\nof x such that x > 15.\nthe mathematical function has an absolute minimum\nat x = 15.\nthe domain of the mathematical function shown in the\ngraph is all real numbers.\nthe average rate of change of the function between x =\n20 and x = 25 is 0.\nthe domain of the mathematical function shown in the\ngraph is all values of x such that 0 ≤ x ≤ 20.

which of the following statements are true about the function graphed here? select all that apply.\nthe mathematical function is increasing for all x such\nthat 15 < x < 22.\nthe mathematical function is decreasing for all values\nof x such that x > 15.\nthe mathematical function has an absolute minimum\nat x = 15.\nthe domain of the mathematical function shown in the\ngraph is all real numbers.\nthe average rate of change of the function between x =\n20 and x = 25 is 0.\nthe domain of the mathematical function shown in the\ngraph is all values of x such that 0 ≤ x ≤ 20.

Answer

Explanation:

Step1: Analyze the function (V(x)=-x^{3}+56x^{2}-1008x + 5760)

First, find the derivative (V^{\prime}(x)=-3x^{2}+112x - 1008). Set (V^{\prime}(x)=0), then (3x^{2}-112x + 1008 = 0). Using the quadratic formula (x=\frac{112\pm\sqrt{112^{2}-4\times3\times1008}}{2\times3}=\frac{112\pm\sqrt{12544 - 12096}}{6}=\frac{112\pm\sqrt{448}}{6}=\frac{112\pm8\sqrt{7}}{6}=\frac{56\pm4\sqrt{7}}{3}\approx\frac{56\pm10.58}{3}). So (x_1=\frac{56 + 10.58}{3}\approx22.19), (x_2=\frac{56-10.58}{3}\approx15.14).

Step2: Check the increasing - decreasing intervals

We can use test points. For (x < 15.14), let (x = 10), (V^{\prime}(10)=-3\times100+112\times10 - 1008=-300 + 1120-1008=-188<0). For (15.14<x<22.19), let (x = 16), (V^{\prime}(16)=-3\times256+112\times16 - 1008=-768+1792 - 1008=16>0). For (x>22.19), let (x = 23), (V^{\prime}(23)=-3\times529+112\times23 - 1008=-1587+2576 - 1008=-219<0).

Step3: Check the average rate of change

The average rate of change formula is (\frac{V(x_2)-V(x_1)}{x_2 - x_1}). (V(20)=-(20)^{3}+56\times(20)^{2}-1008\times20 + 5760=-8000+22400-20160 + 5760=0), (V(25)=-(25)^{3}+56\times(25)^{2}-1008\times25 + 5760=-15625+35000-25200 + 5760=-1065). The average rate of change between (x = 20) and (x = 25) is (\frac{V(25)-V(20)}{25 - 20}=\frac{-1065-0}{5}=-213\neq0).

Step4: Check the domain

Since (x) represents a dimension (from the context of the problem, likely a length - related dimension for the volume of an object), (x\geq0). Also, from the behavior of the function (volume cannot be negative in a practical sense for the given model and the graph), and from the derivative analysis, the relevant domain for the non - negative volume (in the context of the problem's practical interpretation) is (0\leq x\leq20) (because after (x\approx22) the function starts to decrease and may give non - practical negative values for volume in the context of the graphical and problem's implied physical meaning).

Answer:

  • The mathematical function is increasing for all (x) such that (15 < x<22).
  • The domain of the mathematical function shown in the graph is all values of (x) such that (0\leq x\leq20).