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1. The problem statement, all variables and given/known data

Find "b" so that the difference between the x co-ordinates of the two inflection points of y=x^{4}+ 4x^{3}+bx^{2}+5x+7 is 3

3. The attempt at a solution

I.P. at y' = 0

two x co-ordinates are x and x+3

y' = 4x^{3}+ 12x^{2}+ 2bx +5

0 = 4x^{3}+ 12x^{2}+ 2bx +5

0 =4(x+3)^{3}+ 12(x+3)^{2}+ 2b(x+3) +5

4x^{3}+ 12x^{2}+ 2bx +5 = 4(x+3)^{3}+ 12(x+3)^{2}+ 2b(x+3) +5

expand, add, subtract and solve for x, you get:

(-216-6b)/72 = x

I.P is y''=0

y'' = 12x^{2}+ 24x +2b

sub in x

0= 12[(-216-6b)/72]^{2}+24 (-216-6b)/3]+2b

then you expand, add, subtract and end up with the quadratic equation

0= 36 (b^{2}+48b +432)

use the quadratic formula to get b=-12 or or b=-36

to check if b is right, I plugged it back in to y''

0= 12x^{2}+ 24x -2b

0=12x^{2}+ 24x -2(-12)

0 = 12x^{2}+ 24x -24

x=-3.8284 or x=1.8284

but the difference here is 5.6568

0=12x^{2}+ 24x -2(-36)

x= -1, x = 3

the difference here is 4.

Question 2:

1. The problem statement, all variables and given/known data

if f(x) = x^{3}+ 3x^{2}+ k has three distinct real roots, what are the bounds on "k"? (i.e., ? <x<?). Hint: Look for extema using f' and f''.

3. The attempt at a solution

f(x) = x^{3}+ 3x^{2}+ k

f'(x)= 3x^{2}+6x

0= 3x(x+2)

x = 0, x=-2

f"(x) = 6x+6

f'' (0) = 6

at x=0, the function is concave up

f''(-2) = -6

at x = -2, it is concave down

because of the type of graph (its a cubic function), the y-int (which is k) must be equal to or less than 0, but has to be greater than equal to -4...

as I was typing this, I realized that this doesnt make sense, so I dont understand this question.

I will appreciate any help I can get.

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# Homework Help: 2 Calculas Questions

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