Mwaendange M.
asked • 04/27/24find the critical points and discuss the concavity of 5th root x^4 (x-4)^2
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1 Expert Answer
HIRDESHWAR S. answered • 05/06/24
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New to Wyzant
M.Sc. & Ph.D. in Mathematics. Experienc 40 years
Given the function y = x^4/5 (x-4)^2/5
Domain= (-∞, ∞), Range = [0,∞)
y' = dy/dx = (4/5)x-1/5 (x-4)2/5 + (2/5)x4/5(x-4)-3/5 = (2/5)[x-1/5(x-4)-3/5 [2(x-4) + x]
=(2/5) (3x - 8) / [x1/5 (x-4)3/5],
y' = 0 ⇒ 3x - 8 = 0 ⇒ x = 8/3 = 2.667.
At x=2/3, y = (8/3)4/5(8/3 - 4)2/5 = 2.459, Therefore Critical point is at (2.667, 2.459).
y'' = d2y /dx2 = (2/5)[ x1/5 (x-4)3/5(3) - (3x - 8){(1/5) x-4/5 (x-4)3/5 +(3/5)x1/5 (x-4)-2/5}] / [x2/5 (x-4)6/5]
= (2/5) [3x1/5 (x-4)3/5 - (4/5)(3x - 8)x-4/5 (x-4)-2/5 (3x-8)(x-1)] / [x2/5 (x-4)6/5]
= (2/25)[ 3x2-16x-32] / [x6/5 (x - 4)(8/5)]
= 6(x - 6.883)(x+1.55)[25x6/5 (x - 4)(8/5)]
y" (2.667) = -0.83, y is maximum at (2.667, 2.459)
y" >0 for x< -1.55 and x > 6.883 Concave Upwards
y" <0 for -1.55 < x < 6.883, Concave Downwards
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Doug C.
For a similar post from you (4th root of ...), I provided a graph to help you through finding first and 2nd derivative, but that was after you clarified the 4th root of what. Similarly here, it is not clear if the function is the 5th root of the entire [x^4 (x-4)^2] or if it is just the 5th root of x^4 that is intended. You could post as x^(4/5)(x-4)^2 for example.04/28/24