Question

review

1. ( f ( x ) = x ^ { 4 } - x ^ { 3 } - 3 x ^ { 2 } + 1 )

domain: ( ( - infty , infty ) )
range: ( - 4 , infty ) )
degree: 4

of turns: 2

y - int: ( ( - 1,0 ) )
x - int: ( ( - 1,2 ) )
interval of increase:
interval of decrease:
relative max:
relative min:

Explanation:

Step1: Find derivative

The function is \(y = f(x)=x^{4}-x^{3}-3x^{2}+1\). The derivative \(y^\prime=f^\prime(x) = 4x^{3}-3x^{2}-6x=x(4x^{2}-3x - 6)\)

Step2: Find critical points

Set \(y^\prime = 0\), so \(x(4x^{2}-3x - 6)=0\). The solutions of \(4x^{2}-3x - 6=0\) are \(x=\frac{3\pm\sqrt{9+96}}{8}=\frac{3\pm\sqrt{105}}{8}\approx\frac{3\pm10.25}{8}\), and \(x = 0\). The critical points are \(x_1=\frac{3 - \sqrt{105}}{8}\approx - 0.906\), \(x_2 = 0\), \(x_3=\frac{3+\sqrt{105}}{8}\approx1.656\)

Step3: Use first - derivative test

• For \(x\in(-\infty,\frac{3 - \sqrt{105}}{8})\), pick \(x=-1\), \(y^\prime(-1)=4\times(-1)^{3}-3\times(-1)^{2}-6\times(-1)=-4 - 3 + 6=-1<0\), function is decreasing.
• For \(x\in(\frac{3 - \sqrt{105}}{8},0)\), pick \(x =-\frac{1}{2}\), \(y^\prime(-\frac{1}{2})=4\times(-\frac{1}{2})^{3}-3\times(-\frac{1}{2})^{2}-6\times(-\frac{1}{2})=- \frac{1}{2}-\frac{3}{4}+3=\frac{-2 - 3 + 12}{4}=\frac{7}{4}>0\), function is increasing.
• For \(x\in(0,\frac{3+\sqrt{105}}{8})\), pick \(x = 1\), \(y^\prime(1)=4\times1^{3}-3\times1^{2}-6\times1=4 - 3 - 6=-5<0\), function is decreasing.
• For \(x\in(\frac{3+\sqrt{105}}{8},+\infty)\), pick \(x = 2\), \(y^\prime(2)=4\times2^{3}-3\times2^{2}-6\times2=32-12 - 12 = 8>0\), function is increasing.

Step4: Find relative extrema

• Relative maximum: When \(x = 0\), \(y=f(0)=1\), so the relative maximum is \((0,1)\)
• Relative minimum: When \(x=\frac{3 - \sqrt{105}}{8}\), \(y = f(\frac{3 - \sqrt{105}}{8})\approx f(-0.906)=(-0.906)^{4}-(-0.906)^{3}-3\times(-0.906)^{2}+1\approx0.673 + 0.745-2.463 + 1=-0.045\); when \(x=\frac{3+\sqrt{105}}{8}\), \(y=f(\frac{3+\sqrt{105}}{8})\approx f(1.656)=(1.656)^{4}-(1.656)^{3}-3\times(1.656)^{2}+1\approx7.67-4.59-8.19+1=-4.11\)

Step5: Find intervals of increase and decrease

• Interval of increase: \((\frac{3 - \sqrt{105}}{8},0)\cup(\frac{3+\sqrt{105}}{8},+\infty)\)
• Interval of decrease: \((-\infty,\frac{3 - \sqrt{105}}{8})\cup(0,\frac{3+\sqrt{105}}{8})\)

Answer:

• Relative min: \((\frac{3+\sqrt{105}}{8},f(\frac{3+\sqrt{105}}{8}))\approx(1.66,-4.11)\) and \((\frac{3 - \sqrt{105}}{8},f(\frac{3 - \sqrt{105}}{8}))\approx(-0.91,-0.05)\)
• Relative max: \((0,1)\)
• Interval of decrease: \((-\infty,\frac{3 - \sqrt{105}}{8})\cup(0,\frac{3+\sqrt{105}}{8})\)
• Interval of increase: \((\frac{3 - \sqrt{105}}{8},0)\cup(\frac{3+\sqrt{105}}{8},+\infty)\)