Question

help please

locate the critical points and determine the maxima and minima:

1. y=x^4+2x^3+8x+3
2. y=x^3*(x-2)^2

Answer 1

differentiate and equate to zero:

y' = 4x^3 + 6x^2 + 8 = 0

x = -2 is the only real solution

y" = 12x^2 + 12x which is positive at x = -2
so this is a maximum turning point

maximum at x=-2 y = 16 -16 -16 +3 = -13

critical point (maximum) at (-2,-13)

Answer 2

you can solve 2. by the same general method

Answer 3

sorry - i had a mental block !!!! a positive second derivative means a MINIMUM turning point replace maximum by MINIMUM

Answer 4

i'll start number 2 for you

y' = 3x^2 * (x-2)^2 + x^3*2(x-2) = 0

Answer 5

thnx

Answer 6

(x-2)(3x^2(x-2) + 2x^3) = 0
so one solution is x = 2
3x^3-6x^2 + 2x^3 = 0
5x^3 - 6x^2 = 0
x^2(5x - 6 ) = 0
x = 0 or x = 6/5

Answer 7

so there are critical points at x=-2, x=0 and x= 6/5

now we have to find y" and plug in these values to find the nature of these points

Answer 8

y' = (x-2)(5x^3- 6x^2)
= 5x^4 - 6x^3 - 10x^3 + 12x^2
= 5x^4 - 16x^3 + 12x^2
y" = 20x^3 - 48x^2 + 24x
at x=2 this is postive - so a minimum
at x = 6/5 this is negative s0 a minimum
at x = 0 this is also zero - this looks like a point of inflection

MINIMUM AT (2, 2^3(2-2)^2) = (,2,0)
maximum at (1.2, 1.1)
point of inflection at (0,0)

Answer 9

that was a long one!
- i checked it out on graphical calculator to make sure its right

Answer 10

ok - do you follow it alright?

Answer 11

at one point i said x= -2 is one of the points - that was wrong its x = 2 - as i said later

Answer 12

that second one led to some messy algebra

Answer 13

thanks a bunch