y=x^4+bx^2+8x+1 (calculus question)?

Question

y=x^4+bx^2+8x+1 has a horizontal tangent & a point of inflection for the same value of x. What is the value of b?

I took the first derivative. Then I took the second derivative. I don't know where to go from there to find b. Any tips? Thank you in advance.

Answer 1

A horizontal tangent means that the slope (first derivative) at that point is zero.

An inflection point means that the second derivative at that same point is also zero.

So you want to choose b so that the first derivative and the second derivative have a common factor.

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

6x^2 = -b
b = -6x^2

2x^3 - 6x^3 + 4 = 0
-4x^3 = -4
x^3 = 1
x = 1

Therefore b = -6(1) = -6.

Answer 2

y=x^4+bx^2+8x+1

first = 4x^3 + 2bx + 8

second = 12x^2 + 2b

so pt. of inflection when
second = 0

12x^2 + 2b = 0

x = (-2b/12)^0.5

and then tangent when 4x^3 + 2bx + 8 = 0

plug x value in this equation..

would that help?

Answer 3

y=x^4+bx^2+8x+1
y'=4x^3 +2bx +8
y''= 12x^2 +2b

we need a point for which y'=0 and y'' =0
so 4x^3 +2bx +8 =0 and 12x^2 +2b=0
multiply the second eq, by x:
12x^3+2bx=0, => 2bx = -12x^3,

4x^3 -12x^3 + 8=0
-8x^3 +8 =0
x^3 =1, so x=1,
then 2b(1) =-12,
b= -6
.

Answer 4

B is irrellevent to the hypososis. What has incurred is a infringement due to lack of intelligence and big words thrown together.