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Find the maximum points, minimum points, and inflection points, if any in the following functions of x: (a) у %3D — 12х2 + 2х — 8 (b) y =x3 -x² – 30x + 2 (c) у%3 (4х — 4)5 | 2.

Find the maximum points, minimum points, and inflection points, if any in the following functions of x: (a) у %3D — 12х2 + 2х — 8 (b) y =x3 -x² – 30x + 2 (c) у%3 (4х — 4)5 | 2.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Can you please help solve following question 2 a,b and c please?

I attached a similar question with awnsers to a different equation for you perusal if necessary.

Find the maximum points, minimum points, and inflection points, if any in the following functions of x: (a) у %3D — 12х2 + 2х — 8 1 1 y = x³ -x² – 30x + 2 (c) y = (4x – 4)5 2.
Transcribed Image Text:Find the maximum points, minimum points, and inflection points, if any in the following functions of x: (a) у %3D — 12х2 + 2х — 8 1 1 y = x³ -x² – 30x + 2 (c) y = (4x – 4)5 2.
2. Find the maximum points, minimum points, and inflection points, if any in the following functions of x. (a) y = -4x² + 2x – 6 y' = -8x + 2 = 0 x = ! y" = -8 <0 max y =x - x² – 8x + 2 (b) y' = x² – 2x – 8 = 0 IOC (x – 4)(x + 2) = 0 X = 4, x = -2 у" %3 2х — 2 y"(4) = 6 >0 hence local min y"(-2) = -6 < 0 hence local max Check for POI, y" = 0, at x = 1 y" = 2 + 0 hence non stationary POI at x = 1 (c) y = 2(3x – 3)4 This needs the nth derivative test y' = 8(3x – 3)°(3) = 0 at x = 1 y" = 72(3x – 3)²(3) = 0 at x =1 у" %3 432(3х —3) (3) 3 0 at x = 1 yiv = 3888 >0 Differentiated 4 times (even), the derivative is positive so we have a min
Transcribed Image Text:2. Find the maximum points, minimum points, and inflection points, if any in the following functions of x. (a) y = -4x² + 2x – 6 y' = -8x + 2 = 0 x = ! y" = -8 <0 max y =x - x² – 8x + 2 (b) y' = x² – 2x – 8 = 0 IOC (x – 4)(x + 2) = 0 X = 4, x = -2 у" %3 2х — 2 y"(4) = 6 >0 hence local min y"(-2) = -6 < 0 hence local max Check for POI, y" = 0, at x = 1 y" = 2 + 0 hence non stationary POI at x = 1 (c) y = 2(3x – 3)4 This needs the nth derivative test y' = 8(3x – 3)°(3) = 0 at x = 1 y" = 72(3x – 3)²(3) = 0 at x =1 у" %3 432(3х —3) (3) 3 0 at x = 1 yiv = 3888 >0 Differentiated 4 times (even), the derivative is positive so we have a min
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