Question

If f'(x) = Ax^2 + b (b,a are constants)
f(0) + f'(0) = 1 , f(1) = 3
then , f'(1) = .....

Answer 1

I fail to find b :/ the final answer is 4

Answer 2

b = 1

Answer 3

Show me your work.

Answer 4

no!)

Answer 5

i haven't do anything, its only my thoughts

Answer 6

f'(x) = 2Ax

Answer 7

f'(x) = Ax^2 + b.

Answer 8

and f'(0)= 2a*0

Answer 9

ok

Answer 10

just a moment

Answer 11

It's already the first derivative.Take your time.

Answer 12

did you take the integral?

Answer 13

f'(x) =ax^2 + b
f(0) + f'(0) = 1 , f(1) = 3

f(x) = ax^3/3 + bx + c

f(0) = a(0) + b(0) + c
f ' (0) = a(0) + b

Answer 14

I did , The result was 2 unknowns with 1 equation.
A = 6

Answer 15

c + b = 1 (1)
f(1) = 1/3 * a + b + c = 3 from (1)
1 + 1/3 a = 3
so a = 6

Answer 16

correct

Answer 17

f'(1) = a + b = 6 + b. The problem now is to find another equation to find b.

Answer 18

Clarify a bit, I can't understand, well.

Answer 19

so we found that a = 6. lets plug back into original f(x) and f ' (x)

f ' (x) = 6x^2 + b
f (x) = 6/3 x^3 + bx + c = 2x^3 + bx + c

Answer 20

Okie , now ?

Answer 21

hmm, there might be a typo in the directions, im not getting b = 4
can you upload the original directions, take a screen shot

Answer 22

b =-2
a + b = 4

Answer 23

The book is in front of me. I didn't mistype it. "No PC cable or PC Bluetooth"

Answer 24

These are the directions:
If f'(x) = Ax^2 + b (b,a are constants)
f(0) + f'(0) = 1
f(1) = 3

find f ' (1)

Answer 25

That's it.

Answer 26

either the book writer was mistaken or there is magic in it.

Answer 27

right, either its a typo or something else.
what is the name of this book, maybe i can check google book results, the question might pop up

Answer 28

Sorry , I was afk.