hw 16 - first derivative test section 3.2: problem 6 (1 point) consider the function ( f(x)=2x^{2}-6x^{4} )…

hw 16 - first derivative test section 3.2: problem 6 (1 point) consider the function ( f(x)=2x^{2}-6x^{4} ). 1. ( f ) is increasing on the intervals 2. ( f ) is decreasing on the intervals 3. the relative maxima of ( f ) occur at ( x = ) 4. the relative minima of ( f ) occur at ( x = ) (round to four decimal places as needed) notes: use ( inf ) for infinity or ( -inf ) for negative infinity. in the last two, your answer should be a comma separated list of ( x ) values or the word \none\.

hw 16 - first derivative test section 3.2: problem 6 (1 point) consider the function ( f(x)=2x^{2}-6x^{4} ). 1. ( f ) is increasing on the intervals 2. ( f ) is decreasing on the intervals 3. the relative maxima of ( f ) occur at ( x = ) 4. the relative minima of ( f ) occur at ( x = ) (round to four decimal places as needed) notes: use ( inf ) for infinity or ( -inf ) for negative infinity. in the last two, your answer should be a comma separated list of ( x ) values or the word \none\.

Answer

Explanation:

Step1: Find the first derivative

Using the power rule ((x^n)^\prime = nx^{n - 1}), for (y = f(x)=2x^{2}-6x^{4}), the derivative (f^\prime(x)=(2x^{2}-6x^{4})^\prime=4x-24x^{3}=4x(1 - 6x^{2}))

Step2: Find the critical points

Set (f^\prime(x)=0), so (4x(1 - 6x^{2})=0). Solving (4x=0) gives (x = 0), and solving (1-6x^{2}=0) (i.e., (x^{2}=\frac{1}{6})) gives (x=\pm\frac{1}{\sqrt{6}}\approx\pm0.4082)

Step3: Test the intervals

  • For the interval ((-\infty,-\frac{1}{\sqrt{6}})), let (x=- 1). Then (f^\prime(-1)=4\times(-1)-24\times(-1)^{3}=-4 + 24=20>0)
  • For the interval ((-\frac{1}{\sqrt{6}},0)), let (x=-0.2). Then (f^\prime(-0.2)=4\times(-0.2)-24\times(-0.2)^{3}=-0.8+0.192=-0.608<0)
  • For the interval ((0,\frac{1}{\sqrt{6}})), let (x = 0.2). Then (f^\prime(0.2)=4\times0.2-24\times(0.2)^{3}=0.8 - 0.192 = 0.608>0)
  • For the interval ((\frac{1}{\sqrt{6}},\infty)), let (x = 1). Then (f^\prime(1)=4\times1-24\times1^{3}=4 - 24=-20<0)

Since (f^\prime(x)>0) on ((-\infty,-\frac{1}{\sqrt{6}})\cup(0,\frac{1}{\sqrt{6}})), the function (f(x)) is increasing on ((-\infty,-\frac{1}{\sqrt{6}}]\cup[0,\frac{1}{\sqrt{6}}]) Since (f^\prime(x)<0) on ([-\frac{1}{\sqrt{6}},0]\cup[\frac{1}{\sqrt{6}},\infty)), the function (f(x)) is decreasing on ([-\frac{1}{\sqrt{6}},0]\cup[\frac{1}{\sqrt{6}},\infty))

For relative maxima: Using the first - derivative test, when (x=-\frac{1}{\sqrt{6}}) and (x=\frac{1}{\sqrt{6}}), the function changes from increasing to decreasing. So (x=\pm0.4082) are relative maxima. For relative minima: When (x = 0), the function changes from decreasing to increasing. So (x = 0) is a relative minima.

Answer:

  1. ((-\infty,-0.4082]\cup[0,0.4082])
  2. ([-0.4082,0]\cup[0.4082,\infty))
  3. (-0.4082,0.4082)
  4. (0)