2. a. find the critical number(s) of ( q(x)=10x^{11}-11x^{10} ).\nb. where is ( q(x) ) increasing, and where…

Explanation:

Step1: Find the derivative of (q(x))

Using the power rule ((x^n)^\prime=nx^{n - 1}), we have (q^\prime(x)=(10x^{11}-11x^{10})^\prime=10\times11x^{10}-11\times10x^{9}=110x^{10}-110x^{9}=110x^{9}(x - 1))

Step2: Find the critical numbers

Set (q^\prime(x)=0), so (110x^{9}(x - 1)=0). Then (x^{9}=0) or (x - 1=0). Solving these equations gives (x = 0) and (x=1)

Step3: Determine the intervals of increase and decrease

• Choose test points:
  • For (x\lt0), let (x=-1). Then (q^\prime(-1)=110\times(-1)^{9}\times(-1 - 1)=110\times(-1)\times(-2)=220\gt0)
  • For (0\lt x\lt1), let (x=\frac{1}{2}). Then (q^\prime(\frac{1}{2})=110\times(\frac{1}{2})^{9}\times(\frac{1}{2}-1)=110\times\frac{1}{512}\times(-\frac{1}{2})\lt0)
  • For (x\gt1), let (x = 2). Then (q^\prime(2)=110\times2^{9}\times(2 - 1)=110\times512\times1\gt0)
• Intervals:
  • (q(x)) is increasing on ((-\infty,0)\cup(1,\infty))
  • (q(x)) is decreasing on ((0,1))

Step4: Use the First - Derivative Test for local maxima and minima

• At (x = 0):
  • Since (q^\prime(x)) changes sign from positive (when (x\lt0)) to negative (when (0\lt x\lt1)), by the First - Derivative Test, (q(x)) has a local maximum at (x = 0). (q(0)=10\times0^{11}-11\times0^{10}=0)
• At (x = 1):
  • Since (q^\prime(x)) changes sign from negative (when (0\lt x\lt1)) to positive (when (x\gt1)), by the First - Derivative Test, (q(x)) has a local minimum at (x = 1). (q(1)=10\times1^{11}-11\times1^{10}=10 - 11=-1)

Answer:

A. The critical numbers are (x = 0) and (x = 1) B. (q(x)) is increasing on ((-\infty,0)\cup(1,\infty)) and decreasing on ((0,1)) C. Local maximum at ((0,0)) and local minimum at ((1,-1))