Question

1. a. find the critical number(s) of ( q(x)=10x^{11}-11x^{10} ).

b. where is ( q(x) ) increasing, and where is ( q(x) ) decreasing?
c. using the first derivative test, find the local max/min of ( q(x) ).

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 \((1,-1)\)