Question

The function y=x^4+bx^2+8x+1 has a horizontal tangent and a point of inflection for the same value of x. What must be the value of b?

Answer 1

so f ' (x) = 0 and f ' ' (x) = 0 for the same x . so f '(x) = f '' (x)
f ' = 4x^3 + 2bx + 8 = f '' (x) = 12x^2 + 2b

Answer 2

but that still doesn't give me the value of b... do i set it equal to 0?

Answer 3

you have to solve that equation

Answer 4

we have two equations 4x^3+2bx+8=0
and 12x^2+2b=0
I solve the bottom one for b and replace the b in the first equation with -6x^2

Answer 5

4x^3 + 2bx + 8= 12x^2 + 2b

Answer 6

oh thats smarter myininya has it

Answer 7

my way involves a bit of work

Answer 8

if you do this correctly you should get x=1
so now we can find b
since b=-6x^2
then b=-6(1)^2=-6

Answer 9

any questions?

Answer 10

got it thank you!! :)

Answer 11

:)

Answer 12

i'm going to keep posting questions so feel free to answer :) lol... i have my ap test on wednesday so i'm reviewing a lot

Answer 13

good luck

Answer 14

thanks

Answer 15

and you could have done it cantorset's way but it is a bit more thinking down his path

Answer 16

i will show you if you want. i will do it on paper and scan it and post it

Answer 17

I'll try working on it in a bit, but I like the way you taught me better, seems a lot easier.

Answer 18

wow.. looks like a lot of work :)

Answer 19

lol