Question

Find f.
f ''(x) = 24x^2 + 2x + 6, f(1) = 2, f '(1) = −3

I know f''(x) = 2x^4+(x^3/3)+3x^2

Answer 1

Did you mean to say f(x) = 2x^4+(x^3/3)+3x^2 ?

Answer 2

yes opps

Answer 3

you integrated properly, but you forgot about the +C both times

Answer 4

if f ''(x) = 24x^2 + 2x + 6, then what is f'(x) ?

Answer 5

refresh the page if you get weird symbols showing up

Answer 6

f'(x)=8x^3+x^2+6x+C

Answer 7

correct

Answer 8

now use the fact that f'(1) = -3

Answer 9

to find the value of C in f'(x)=8x^3+x^2+6x+C

Answer 10

Like the following?

1=8(-3)^3+(-3)^2+6(-3)+C

Answer 11

the other way around

Answer 12

x = 1, f ' (x) = -3

Answer 13

Okay, I got C=-17

Answer 14

close

Answer 15

-18

Answer 16

correct

Answer 17

therefore, f ' (x) = 8x^3+x^2+6x-18

Answer 18

do I do the anti der of that?

Answer 19

yes

Answer 20

okay, I got f(x)= 2x^4 + (x^3)/3 + 3x^2 - 18x + c

Answer 21

me too

Answer 22

now use f(1) = 2 to find C

Answer 23

C=44/3

Answer 24

correct

Answer 25

So I'm assuming the final answer would be f(x)= 2x^4 + (x^3)/3 + 3x^2 - 18x + 44/3

Answer 26

yes it is

Answer 27

thanks!

Answer 28

you're welcome