Solved: Find "k" for Distinct Real Roots of f(x) = x3 + 3x2 + k

[[ScPpL]Shree]

Question 1:

Homework Statement

Find "b" so that the difference between the x co-ordinates of the two inflection points of y=x⁴ + 4x³+bx²+5x+7 is 3

The Attempt at a Solution

I.P. at y' = 0
two x co-ordinates are x and x+3
y' = 4x³ + 12x² + 2bx +5
0 = 4x³ + 12x² + 2bx +5
0 =4(x+3)³ + 12(x+3)² + 2b(x+3) +5

4x³ + 12x² + 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'' = 12x² + 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 (b²+48b +432)

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

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

0= 12x² + 24x -2b
0=12x² + 24x -2(-12)
0 = 12x² + 24x -24
x=-3.8284 or x=1.8284
but the difference here is 5.6568

0=12x² + 24x -2(-36)
x= -1, x = 3
the difference here is 4.

Question 2:

Homework Statement

if f(x) = x³ + 3x² + k has three distinct real roots, what are the bounds on "k"? (i.e., ? <x<?). Hint: Look for extema using f' and f''.

The Attempt at a Solution

f(x) = x³ + 3x² + k
f'(x)= 3x² +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 doesn't make sense, so I don't understand this question.

I will appreciate any help I can get.

 

[ehild]

Question 1:

Find "b" so that the difference between the x co-ordinates of the two inflection points of y=x⁴ + 4x³+bx²+5x+7 is 3

The Attempt at a Solution

I.P. at y' = 0
two x co-ordinates are x and x+3

The inflection points means y''=0.

Question 2:if f(x) = x³ + 3x² + k has three distinct real roots, what are the bounds on "k"? (i.e., ? <x<?). Hint: Look for extema using f' and f''.

The Attempt at a Solution

f(x) = x³ + 3x² + k
f'(x)= 3x² +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 doesn't make sense, so I don't understand this question.
.

The result is almost good, but I do not understand the explanation. Sketch the graph. It has three different real roots so it has three x points where y change from negative to positive or vice versa. It has two extrema, one at x=-2, one at x =0. What are these extrema, where is minimum and where is maximum? Where should be the function positive and where is it negative?

ehild