How do you find all local maximum and minimum points using the second derivative test given #y=x^5-x#?

Question

2444 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-23T02:00:38

#y_max = y(approx-0.669) approx 0.535#
#y_min = y(approx+0.669) approx -0.535#

#y = x^5-x#

#dy/dx = 5x^4-1#

#(d^2y)/dx^2 = 20x#

For local extrema #dy/dx=0#

#:. 5x^4-1 =0#

#5x^4 =1#

#x^4 =1/5#

#x = +-(1/5)^(1/4), +-(1/5)^(1/4)i#

Since we are only interested in real roots

#x=+-(1/5)^(1/4) approx +-0.669#

For a local maximum #(d^2y)/dx^2 <0#

For a local minimum #(d^2y)/dx^2 >0#

Testing our results from above

#xapprox+0.669 -> 20xx x >0 -> y(approx+0.669) = y_min#

#xapprox-0.669 -> 20xx x <0 -> y(approx-0.669) = y_max#

Hence:

#y_max = y(approx-0.669) approx 0.535#
#y_min = y(approx+0.669) approx -0.535#

We can see the local maximum and minimum on the graph of #y# below.

graph{x^5-x [-2.434, 2.434, -1.217, 1.217]}