Question

solve f(x)=2 by the function f(x)=x^3-3x^2+3x

Answer 1

This problem is asking you what x is when f(x) is 2.
Start by setting your equation equal to 2.

Answer 2

solve x^3-3x^2+3x-2=0

Answer 3

solved it.so the formula was x^3-3x^2+3x
x=3x^2-6x=3x
f(x)=6x-6+3
f'(x)=6x-6=0+3
0+3=3
thus,f'(x)=3

Answer 4

there may have been a bit of info i left out.it was a table with all the x,f(x) and f'(x) values.

Answer 5

i had to solve f'(x) when f(x)=2 while using the function mentioned.

Answer 6

hey,guys,could you help me out with understanding the question?

Answer 7

Hey @nck93 we are given f(x)=2 and f(x) is defined as x^3-3x^2+3x
so
\[f(x)=x^3-3x^2+3x\]
and for a particular value of x f(x) is 2

Answer 8

yeah,i get that.i understand solving x and f(x), but not f'(x).like in the attachment the value for f'(x) is not given,but the value for for f(x) is 2 the same on the left side,but with 0,how do i solve that?

Answer 9

Wait I'm going through attachement

Answer 10

x=0 we have
\[ f(x)=x^3-4x^2+3x\]
put x=0
we get
\[f(0)=0\]
that's why table mentions f(x)=0 for x=0
now differentiate with respect to x
\[f'(x)=3x^2-8x+3\]
put x=0 to find f'(x) at x=0

Answer 11

ah,now i get it.thanks alot =)

Answer 12

welcome:)

Answer 13

im using 3x^2-8x+3 as 3(0)^2-8(0)+3 is that right?

Answer 14

Yeah:)

Answer 15

sorry 3(0) i think you noted the question down wrong. its +3x in the end,not 3.

Answer 16

oh wait,my mistake.that after differentiation..

Answer 17

its f(x)= x^3-3X^2+3x you wrote x^3-4x^2+3x

Answer 18

Sorry
\[f'(x)=3x^2-6x+3\]
now put x=0 to find f'(x) at x=0

Answer 19

Np,thanks.

Answer 20

welcome:)