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

Question

Christopher Allers · Accepted Answer

#y_max = y(approx-0.669) approx 0.535#
#y_min = y(approx+0.669) approx -0.535# #y = x^5-x# Apply power rule #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]}

Evelyn Agard · Answer

undefined To find all local maximum and minimum points using the second derivative test for the function $ y = x^5 - x $, follow these steps: 1. Find the critical points by setting the first derivative equal to zero and solving for $ x $. $$ \frac{dy}{dx} = 5x^4 - 1 = 0 $$ $$ 5x^4 = 1 $$ $$ x^4 = \frac{1}{5} $$ $$ x = \pm \sqrt[4]{\frac{1}{5}} $$ 2. Find the second derivative $ \frac{d^2y}{dx^2} $. $$ \frac{d^2y}{dx^2} = 20x^3 $$ 3. Evaluate the second derivative at each critical point to determine the nature of the stationary points: - If $ \frac{d^2y}{dx^2} > 0 $, the point is a local minimum. - If $ \frac{d^2y}{dx^2} < 0 $, the point is a local maximum. - If $ \frac{d^2y}{dx^2} = 0 $, the test is inconclusive. 4. Substitute the critical points into the second derivative: - For $ x = -\sqrt[4]{\frac{1}{5}} $: $ \frac{d^2y}{dx^2} = 20(-\sqrt[4]{\frac{1}{5}})^3 < 0 $, so it is a local maximum. - For $ x = \sqrt[4]{\frac{1}{5}} $: $ \frac{d^2y}{dx^2} = 20(\sqrt[4]{\frac{1}{5}})^3 > 0 $, so it is a local minimum.