Question

let f(x)=ax^3+6x^2+bx+4. Determine the constant a and b such that f has a relative min at x=-1 and a rel max at x=2@Algebra

Answer 1

I hate these

Answer 2

first derivative of f(x) first

Answer 3

I did that then what?
I got F'(x)=3ax^2+12x+b

Answer 4

\[f'(x)=3a^2+12x+b\]
\[f'(-1)=0 ;f'(2)=0\]

Answer 5

\[f'(x)=3ax^2+12x+b*\]

Answer 6

I was just about to say....what happened to the x?

Answer 7

\[f'(-1)=3a-12+b=0\]

Answer 8

\[f'(2)=3a(4)+12(2)+b=0\]

Answer 9

\[3a+b=12\]
\[12a+b=-24\]

Answer 10

3a + b =12
12a + 24 +b = 0
12a + b =-24
-(3a + b) = 12
9a = -36
a = -4
b =24

Answer 11

what happened to the x?

Answer 12

\[ 3a+b=12\]
\[-(12a+b=-24)\]
------------------
-9a+0b=36
-9a=36
a=-4
-12+b=12
b=24
gj money

Answer 13

what do you mean what happen to x?

Answer 14

how did you go from f'(x)=3ax^2+12x+b* to f′(−1)=3a−12+b=0

Answer 15

\[f(a)=-4x^3+6x^2+24x+4\]

Answer 16

-1 and 2 are critical numbers
this is a polynomial
so our critical numbers only exist when f'=0

Answer 17

f'(-1)=0
f'(2)=0

Answer 18

this was given to us

Answer 19

money and i applied these conditions to f'

Answer 20

I got it. Thank you so much!

Answer 21

here b can take any value and a is rrestricted to any value in the interval (-2,2)