Question

If you have
f(1) = 5
f(3) = 21

How can you find f'(3)?

Answer 1

that's all you're given?

Answer 2

Oh sorry i didnt see the other part on the next page O:

For all real numbers of a and b, f(a+b) - f(a) = kab+2b^2

Answer 3

A= 1 and B= 2

Answer 4

ok figured that much, but what about k

Answer 5

oh nvm, one sec

Answer 6

f(a+b) - f(a) = kab+2b^2

f(1+2) - f(1) = k*1*2+2*2^2

f(3) - f(1) = k*1*2+2*2^2

21 - 5 = 2k - 8

solve for k to get k = ???

Answer 7

4

Answer 8

no

Answer 9

try again

Answer 10

But you put minus 8 when it's positive O:

Answer 11

oh my bad

Answer 12

21 - 5 = 2k + 8

yeah k = 4

Answer 13

so

f(a+b) - f(a) = kab+2b^2

becomes

f(a+b) - f(a) = 4ab+2b^2

Answer 14

now if you factor out b from the right side, you will get this

f(a+b) - f(a) = b(4a + 2b)

then divide both sides by b to get

[ f(a+b) - f(a) ]/b = 4a + 2b

Answer 15

now if you take the limit of both sides as b ---> 0, then you will get the derivative f ' (a)

replace a with 3 to get f ' (3)