Question

Calculus?
apply the mean value theorem to f on [0,1] to find all values of c such that f'(c)= f(b)-f(a)/b-a.

f(x)=x^3

Answer 1

start by evaluating f(a), f(b) and f'(x)

Answer 2

f(a) = ?
f(b) = ?
f'(x) = ?

Answer 3

f'(x)=3x^2

Answer 4

a is 0 and b is 1 right?

Answer 5

yup! a and b are just the boundaries in given interval

Answer 6

f'(x)=3x^2

f'(c) = 3c^2

Answer 7

a = 0, b = 1

f(0) = 0^3 = 0
f(1) = 1^3 = 1

Answer 8

yeah

Answer 9

f'(c)= f(b)-f(a)/b-a

plug them above

Answer 10

3c^2 = 1-0/1-0

3c^2 = 1/1

3c^2 = 1

c^2 = 1/3

c = ?

Answer 11

c=0.576 ?

Answer 12

cause you square both sides right?

Answer 13

c = 1/sqrt(3) = 0.576 is \(\large \color{red}{\checkmark}\)

Answer 14

notice that we discarded the negative value -1/sqrt(3) cuz, it doesnt lie in given interval : [0, 1]

Answer 15

got it

Answer 16

again thank you so much

Answer 17

hey wait a sec, we did a mistake

Answer 18

where?

Answer 19

nope, it looks perfect ! ok

Answer 20

;)

Answer 21

lol thank you ;)

Answer 22

np, u wlc :)