Help with unit 2 Calculus please!

[sally baker]

dy/dx and the answer at the book of the book says 4x-1/3

[b^3]

dy/dx and the answer at the book of the book says 4x-1/3

By calc, 2x^2 +3 is right, so maybe the book is wrong?

[sally baker]

okay thanks and for the questions . find the coordinates of the points on the curves given by the following equations at which the gradient has the given value .

y= x^2-3x+1 , dy/dx =0

okay so can i solve for x in the equation except i get a weird answer  . Im confused :/

[brightsky]

dy/dx = 2x - 3 = 0
x = 3/2, plug back in, y = (3/2)^2 - 3(3/2) + 1 = 9/4 - 9/2 + 1 = -5/4
hopefully arithmetic's not dodgy

[sally baker]

thanks

[b^3]

dy/dx = 2x - 3 = 0
x = 3/2, plug back in, y = (3/2)^2 - 3(3/2) + 1 = 9/4 - 9/2 + 1 = -5/4
hopefully arithmetic's not dodgy

Yep, it's good.

[sally baker]

for the function with rule f(x)=3(2x-1)^2 find the values of x for which

f(x)=0
i know i do the equation and say =0 however i aint getting x correct for all these questions. im so gonna fail this calculus test

[sally baker]

and another question please . sorry . 

the curve with equation y=ax^2+bx has a gradient of 3 at the point (1,1) . find

a)values of a and b                    b) the coordinates of the points where the gradient is 0

[vea]

for the function with rule f(x)=3(2x-1)^2 find the values of x for which

f(x)=0
i know i do the equation and say =0 however i aint getting x correct for all these questions. im so gonna fail this calculus test

f(x)=0
3(2x-1)^2=0
(2x-1)^2=0
2x-1=0
2x=1
x=1/2

and another question please . sorry . 

the curve with equation y=ax^2+bx has a gradient of 3 at the point (1,1) . find

a)values of a and b                    b) the coordinates of the points where the gradient is 0

a) dy/dx=2ax+b
Since dy/dx=3 at x=1,
3=2a+b
b=3-2a...(1)
Now we also know that the point (1,1) is on the graph of y so we can sub y=1 when x=1.
1=a+b
b=1-a...(2)
sub (1) into (2)
3-2a=1-a
a=2, sub into (2)
b=-1
Therefore a=2 and b=-1 and y=2x^2-1

b)When dy/dx=0
2ax+b=0
4x-1=0 (since we have a=2 and b=-1)
x=1/4
When x=1/4, y=1/8-1=-7/8
Therefore the co-ordinates of the point where the gradient is 0 is (1/4, -7/8)

Hopefully I haven't made any mistakes! :S

[REBORN]

and another question please . sorry . 

the curve with equation y=ax^2+bx has a gradient of 3 at the point (1,1) . find

a)values of a and b                    b) the coordinates of the points where the gradient is 0

dy/dx = 2ax + b

When x = 1 the gradient = 3

3 = 2a + b

2nd equation is from subbing the point into 'y'

1= a + b

Then solve simultaneously

EDIT: Beaten

[sally baker]

how to u factorise X^3-2x . i forgot okay LOL

[Greatness]

x^3 - 2x = x(x^2-2) Take x out as a common factor

[sally baker]

okay how would i do the question.  a) find the equation of the tangent at the point (2,4) to the curve y=x^3-2x    i did i got y=10x-16  which is correct

i need help with part b

find the coordinates of the point where the tangent meets the curve again .

[sally baker]

thanks alot.

8 a) find the equation of the tangent to the curve y=x^3-9x^2+20x-8 at the point(1,4)
b) at what points of the curve is the tangent parallel to the line 4x+y-3=0

[nbhindi]

Let f(x)=y=x^3-2x
f'(x)= 3x^2 -2

now from part a) u know the tangent has gradient of 10---> .: let f'(x)=10, solve for x to get x=2, -2

Hence the x value of the point where the tangent meets the curve again is -2
Sub x= -2 into f(x)---->f(-2)=-4
.: coordinates =(-2,-4)