# Math Help - Calculus Graph

1. ## Calculus Graph

Clicky Clicky

I am unable to determine the co-ord of C... well I get a value but it doesnt look right, I get $(1 ; 0)$ which does not seem to fit the value of C. All the other values seem to be perfect execpt for C. Here are my workings.

14.1) $f(x) = (x-2)^2(2x-1)=0$
$(x^2-4x+4)(2x-1)=0$
$2x^3-9x^2+12x-4=0$
$A = y-intercept$
$y=-4 (0 ; -4)$

$B = (2x-1)$
$x = 1/2$
$(1/2 ; 0) = B$

$C = f(x) = 2x^3-9x^2+12-4$
$f'(x) = 6x^2-18+12
x^2-3x+2=0$

$(x-1)(x-2)=0$
$f(1) = 6(1)^2-18(1) +12 = 0$
$C=( 1; 0)$

$f(2) = 6(2)^2-18(2)+12=0$
$(2;0) = D$ Which can also be taken from $(x-2)$

Either the book is incorrect or I am. The book has been incorrect a few times so ye :/

2. To find C you note that it is a turning point.
Thus,
$f'(x)=0$
Since,
$f(x)=(x-2)^2(2x-1)$
The derivative is, (product rule)
$f'(x)=((x-2)^2)'(2x-1)+(x-2)^2(2x-1)'$
Use the chain rule now,
$f'(x)=2(x-2)(2x-1)+2(x-2)^2$
Factor,
$f'(x)=2(x-2)[(2x-1)+(x-2)]$
Simplfy,
$f'(x)=2(x-2)(3x-3)$
You can factor again,
$f'(x)=6(x-2)(x-1)$
Set equal to zero,
$6(x-2)(x-1)=0$
Divide by 6,
$(x-2)(x-1)=0$
Each factor equal to zero,
$x-2=0\rightarrow x=2$
$x-1=0\rightarrow x=1$
One of these points is C and one is D, but which one is it?
Note C is to the left of D thus C<D.
Thus,
$x=1$

3. Thanks. I havent seen that way of solving it like that. I agree that C is 1 but look where it is on the graph. But how can C be (1;0) look where it is on the grapg. That pos means that it has a x and y value. I am confused about that

4. I would also like to know how to solve for y PH
do you just substitute x=1 into the original equation?

5. Originally Posted by Quick
I would also like to know how to solve for y PH
do you just substitute x=1 into the original equation?
Yes,
so the point is
$(1,1)$, not what you said NeF, $(1,0)$ that is wrong.
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I think what your problem is. You are thinking since it is a turning point it is zero. Wrong! The derivative is zero, not necessarily the point is zero.

6. thanx for clearing that up