methods unit 2 help

[sally baker]

can someone help sketch the graphs of each of the following  y=2+3x-x^3  given the type of stationary point  and  represent it by that table you know were your write the  x values one row then the F'(x) then the shape of f .

[b^3]

Find the stationary point, find f'(x) and let it equal 0
f'(x)=3-3x²
3-3x²=0
3=3x²
x²=1
x=plus/minus 1
So find the points
f(1)=2+3-1=4
(1,4)
f(-1)=2-3+1=0
(-1,0)
Now I think you are talking about the gradient table.
What we do is pick the points of the x values (at stationary points) and then x values either side of them and find the gradient, allowing us to determine the nature of the turning point.
So we have  (red will be the t.p.s)
x                -2     -1     0     1     2
m=f'(x)       -9      0     3     0     -9
                                   (1,4)
                   \                 __
                    \     __    /       \
                        (-1,0)           \
So (-1,0) is a minimum and (1,4) is a maximum.
I hope that helps.

EDIT: Just note, use f'(x) for the derivative, not F'(x) as the captial F is sometimes used to denote and antiderivative.

[sally baker]

thankyou so much.  btw  how would i solve x for this equation y=x^3-6x^2+9x+10

[b^3]

thankyou so much.  btw  how would i solve x for this equation y=x^3-6x^2+9x+10

I think that one has to be done by cacl, since it is not easily factorised.

[sally baker]

yeah calculator does some negative decimals maybe using factor theorem

[b^3]

yeah calculator does some negative decimals maybe using factor theorem

Yeh thats the only solution, factor theorem won't work, graph it and you will see what I mean.

[sally baker]

okay well this is what the question says . for the function y=x^3-6x^2+9x+10

a)find the intervals where y is increasing ie( dy/dx >0 )
b)find the stationary points on the curve corresponding to y=x^3-6x^2+9x+10

[b^3]

first you need to find dy/dx
so dy/dx = 3x²-12x+9
So to find where it is increasing you need to find where the derivative is greater than 0.
So first find where it is equal to 0
so  3x²-12x+9=0
3(x-3)(x-1)=0
x=1,3
Now where it is increasing, dy/dx will have a positive value, so try values between these points.
f'(0)=9 (+ve)
f'(2)=12-24+9=-3 (-ve)
f'(4)=48-48+9=9 (+ve)
so f(x) will be increasing for (-infinty,1)U(3,infinity)
b) kinda answered it in a, just find the points (i.e. find y to match x)
f(1)=1-6+9+10
=14
(1,14)
f(3)=27-54+27+10
=10
(3,10)
hopefully there are no mistakes there.

EDIT: So what you are really doing is finding where the gradient is 0 so that you can split the graph of y up and then find where the gradient is postive or negative. If it is positive, y is increasing as x increase, if it is ive y decreases as x increases. Just note, you are not trying to find the x-intercepts of y.